100 results found for "coterminal-equation" in Class 10.
यदि किसी द्विघात समीकरण की जड़ें \(2+\sqrt{5}\) और \(2-\sqrt{5}\) हैं, तो समीकरण कौन-सा है?
If the roots of a quadratic equation are \(2+\sqrt{5}\) and \(2-\sqrt{5}\), which is the equation?
#quadratic-roots
#forming-equation
#surd-roots
A \(x^2-4x-1=0\)
B \(x^2+4x-1=0\)
C \(x^2-4x+1=0\)
D \(x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x-1=0\)
Step 1
Concept
\(The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-4x-1=0). The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 3
Exam Tip
जड़ों का योग (4) और गुणनफल (-1) है। \(समीकरण (x^2-(\)योग)x+गुणनफल\(=0) से उत्तर मिलता है\)।
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यदि किसी द्विघात समीकरण की जड़ें एक-दूसरे की व्युत्क्रम हैं और उनका योग \(\frac{5}{2}\) है, तो समीकरण कौन-सा हो सकता है?
If the roots of a quadratic equation are reciprocals of each other and their sum is \(\frac{5}{2}\), which equation is possible?
#quadratic-roots
#reciprocal-roots
#forming-equation
A \(2x^2-5x+2=0\)
B \(2x^2+5x+2=0\)
C \(x^2-5x+2=0\)
D \(5x^2-2x+2=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-5x+2=0\)
Step 1
Concept
Reciprocal roots have product (1). Multiplying \(x^2-\frac{5}{2}x+1=0\) by (2) gives \(2x^2-5x+2=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-5x+2=0\). Reciprocal roots have product (1). Multiplying \(x^2-\frac{5}{2}x+1=0\) by (2) gives \(2x^2-5x+2=0\).
Step 3
Exam Tip
व्युत्क्रम जड़ों का गुणनफल (1) होता है। समीकरण \(x^2-\frac{5}{2}x+1=0\) को (2) से गुणा करने पर \(2x^2-5x+2=0\) मिलता है।
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यदि किसी द्विघात समीकरण की जड़ों का योग (7) और गुणनफल (10) है, तो समीकरण कौन-सा है?
If the sum of roots of a quadratic equation is (7) and the product is (10), which is the equation?
#quadratic-roots
#forming-equation
#sum-product
A \(x^2+7x+10=0\)
B \(x^2-10x+7=0\)
C \(x^2-7x+10=0\)
D \(x^2+10x-7=0\)
Explanation opens after your attempt
Correct Answer
C. \(x^2-7x+10=0\)
Step 1
Concept
\(The equation is (x^2-(\)sum)x+product\(=0). Hence (x^2-7x+10=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is C. (x^2-7x+10=0). The equation is (x^2-(\)sum)x+product\(=0). Hence (x^2-7x+10=0) is correct.\)
Step 3
Exam Tip
\(जड़ों का समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-7x+10=0) सही है\)।
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यदि किसी द्विघात समीकरण की जड़ें (3) और (-2) हैं और \(x^2\) का गुणांक (2) है, तो समीकरण कौन-सा है?
If the roots of a quadratic equation are (3) and (-2), and the coefficient of \(x^2\) is (2), which is the equation?
#quadratic-roots
#forming-equation
#leading-coefficient
A \(2x^2-2x-12=0\)
B \(2x^2+2x-12=0\)
C \(2x^2-2x+12=0\)
D \(x^2-x-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-2x-12=0\)
Step 1
Concept
The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-2x-12=0\). The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 3
Exam Tip
मॉनिक समीकरण \(x^2-x-6=0\) है। \(x^2\) का गुणांक (2) चाहिए, इसलिए पूरे समीकरण को (2) से गुणा करें।
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यदि किसी मोनिक द्विघात समीकरण के दोनों मूल (5) और (5) हैं तो समीकरण कौन सा है?
If both roots of a monic quadratic equation are (5) and (5), which equation is it?
#roots
#equal_roots
#equation_from_roots
A \(x^2-10x+25=0\)
B \(x^2+10x+25=0\)
C \(x^2-5x+25=0\)
D \(x^2+5x-25=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-10x+25=0\)
Step 1
Concept
If both roots are (5), the equation is ((x-5)2 =0). Expanding it gives \(x^2-10x+25=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+25=0\). If both roots are (5), the equation is ((x-5)2 =0). Expanding it gives \(x^2-10x+25=0\).
Step 3
Exam Tip
दोनों मूल (5) हों तो समीकरण ((x-5)2 =0) होगा। इसे खोलने पर \(x^2-10x+25=0\) मिलता है।
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यदि किसी मोनिक द्विघात समीकरण के दोनों मूल (-4) और (-4) हैं तो समीकरण कौन सा है?
If both roots of a monic quadratic equation are (-4) and (-4), which equation is it?
#roots
#equal_roots
#equation_from_roots
A \(x^2+8x+16=0\)
B \(x^2-8x+16=0\)
C \(x^2+4x+16=0\)
D \(x^2-4x-16=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+8x+16=0\)
Step 1
Concept
If both roots are (-4), the equation is ((x+4)2 =0). Expanding it gives \(x^2+8x+16=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+8x+16=0\). If both roots are (-4), the equation is ((x+4)2 =0). Expanding it gives \(x^2+8x+16=0\).
Step 3
Exam Tip
दोनों मूल (-4) हों तो समीकरण ((x+4)2 =0) होगा। इसे खोलने पर \(x^2+8x+16=0\) मिलता है।
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यदि किसी मोनिक द्विघात समीकरण के दोनों मूल (7) और (7) हैं तो समीकरण कौन सा है?
If both roots of a monic quadratic equation are (7) and (7) then which equation is it?
#roots
#equal_roots
#equation_from_roots
A \(x^2-14x+49=0\)
B \(x^2+14x+49=0\)
C \(x^2-7x+49=0\)
D \(x^2+7x-49=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-14x+49=0\)
Step 1
Concept
With both roots (7) we get ((x-7)2 =0) which is \(x^2-14x+49=0\). Form a perfect square from repeated roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-14x+49=0\). With both roots (7) we get ((x-7)2 =0) which is \(x^2-14x+49=0\). Form a perfect square from repeated roots.
Step 3
Exam Tip
दोनों मूल (7) होने पर ((x-7)2 =0) मिलता है जो \(x^2-14x+49=0\) है। दोहराए मूल से पूर्ण वर्ग बनाएं।
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कंकाली समीकरण और संतुलित समीकरण में मुख्य अंतर क्या है?
What is the main difference between a skeletal equation and a balanced equation?
#science
#class10
#expert
#skeletal-equation
#balanced-equation
A कंकाली समीकरण में परमाणु असमान हो सकते हैं पर संतुलित में समान होते हैं / Atoms may be unequal in skeletal equation but equal in balanced equation
B कंकाली समीकरण में अभिकारक नहीं होते / Skeletal equation has no reactants
C संतुलित समीकरण में उत्पाद नहीं होते / Balanced equation has no products
D दोनों में कोई अंतर नहीं होता / There is no difference between them
Explanation opens after your attempt
Correct Answer
A. कंकाली समीकरण में परमाणु असमान हो सकते हैं पर संतुलित में समान होते हैं / Atoms may be unequal in skeletal equation but equal in balanced equation
Step 1
Concept
A skeletal equation gives only the framework of formulae.
Step 2
Why this answer is correct
Atoms may not be equal in it.
Step 3
Exam Tip
In a balanced equation every element is equal on both sides. चरण 1: कंकाली समीकरण केवल सूत्रों का ढांचा देता है। चरण 2: उसमें परमाणु बराबर न भी हों तो संभव है। चरण 3: संतुलित समीकरण में हर तत्व दोनों ओर बराबर होता है।
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कंकाली समीकरण और संतुलित समीकरण में मुख्य अंतर क्या है?
What is the main difference between a skeletal equation and a balanced equation?
#science
#class10
#medium
#skeletal-equation
#balanced-equation
A कंकाली समीकरण में परमाणु बराबर नहीं हो सकते पर संतुलित समीकरण में बराबर होते हैं / Atoms may not be equal in a skeletal equation but are equal in a balanced equation
B कंकाली समीकरण में कोई पदार्थ नहीं होता / A skeletal equation has no substance
C संतुलित समीकरण में कोई उत्पाद नहीं होता / A balanced equation has no product
D दोनों में कोई अंतर नहीं होता / There is no difference between them
Explanation opens after your attempt
Correct Answer
A. कंकाली समीकरण में परमाणु बराबर नहीं हो सकते पर संतुलित समीकरण में बराबर होते हैं / Atoms may not be equal in a skeletal equation but are equal in a balanced equation
Step 1
Concept
A skeletal equation gives only the formula framework.
Step 2
Why this answer is correct
A balanced equation has equal atoms of each element on both sides.
Step 3
Exam Tip
Therefore balancing makes it complete. चरण 1: कंकाली समीकरण केवल सूत्रों का ढांचा देता है। चरण 2: संतुलित समीकरण में हर तत्व के परमाणु दोनों ओर बराबर होते हैं। चरण 3: इसलिए संतुलन इसे पूर्ण बनाता है।
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शब्द समीकरण की तुलना में रासायनिक समीकरण अधिक उपयोगी क्यों होता है?
Why is a chemical equation more useful than a word equation?
#science
#class10
#word-equation
#chemical-equation
A यह पदार्थों के सूत्र और अनुपात दिखाता है / It shows formulae and ratios of substances
B यह केवल रंग दिखाता है / It shows only colour
C यह केवल पात्र दिखाता है / It shows only container
D यह केवल गंध बताता है / It tells only smell
Explanation opens after your attempt
Correct Answer
A. यह पदार्थों के सूत्र और अनुपात दिखाता है / It shows formulae and ratios of substances
Step 1
Concept
A word equation gives names of substances.
Step 2
Why this answer is correct
A chemical equation shows formulae and balancing.
Step 3
Exam Tip
So it is more precise in exams. चरण 1: शब्द समीकरण पदार्थों के नाम बताता है। चरण 2: रासायनिक समीकरण पदार्थों के सूत्र और संतुलन को दिखाता है। चरण 3: इसलिए परीक्षा में रासायनिक समीकरण अधिक स्पष्ट माना जाता है।
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यदि (4) और (9) किसी द्विघात समीकरण के मूल हैं, तो वह समीकरण कौनसा हो सकता है?
If (4) and (9) are roots of a quadratic equation, which equation can it be?
#quadratic
#construct-equation
#roots
A \(x^2-13x+36=0\)
B \(x^2+13x+36=0\)
C \(x^2-36x+13=0\)
D \(x^2+36x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+36=0\)
Step 1
Concept
If roots are (4) and (9), then ((x-4)(x-9)=0), that is \(x^2-13x+36=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+36=0\). If roots are (4) and (9), then ((x-4)(x-9)=0), that is \(x^2-13x+36=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (4) और (9) हों तो ((x-4)(x-9)=0), यानी \(x^2-13x+36=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
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यदि (3) और (7) किसी द्विघात समीकरण के मूल हैं, तो वह समीकरण कौनसा हो सकता है?
If (3) and (7) are roots of a quadratic equation, which equation can it be?
#quadratic
#construct-equation
#roots
A \(x^2-10x+21=0\)
B \(x^2+10x+21=0\)
C \(x^2-21x+10=0\)
D \(x^2+21x+10=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-10x+21=0\)
Step 1
Concept
If roots are (3) and (7), then ((x-3)(x-7)=0), that is \(x^2-10x+21=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+21=0\). If roots are (3) and (7), then ((x-3)(x-7)=0), that is \(x^2-10x+21=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (3) और (7) हों तो ((x-3)(x-7)=0), यानी \(x^2-10x+21=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
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यदि (2) और (5) किसी द्विघात समीकरण के मूल हैं, तो वह समीकरण कौनसा हो सकता है?
If (2) and (5) are roots of a quadratic equation, which equation can it be?
#quadratic
#construct-equation
#roots
A \(x^2-7x+10=0\)
B \(x^2+7x+10=0\)
C \(x^2-10x+7=0\)
D \(x^2+10x+7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is \(x^2-7x+10=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+10=0\). If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is \(x^2-7x+10=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (2) और (5) हों तो समीकरण ((x-2)(x-5)=0) यानी \(x^2-7x+10=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
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यदि \(x^2-7x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(2\alpha+3\) और \(2\beta+3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-7x+12=0\), which equation has roots \(2\alpha+3\) and \(2\beta+3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-20x+99=0\)
B \(x^2-14x+99=0\)
C \(x^2-20x+91=0\)
D \(x^2+20x+99=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-20x+99=0\)
Step 1
Concept
The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+99=0\). The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 3
Exam Tip
मूल जड़ें (3) और (4) हैं। नई जड़ें (9) और (11) हैं, इसलिए समीकरण \(x^2-20x+99=0\) है।
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यदि \(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha+\beta\) और \(\alpha\beta\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are roots of \(x^2-5x+6=0\), which equation has roots \(\alpha+\beta\) and \(\alpha\beta\)?
#quadratic-roots
#new-equation
#sum-product-roots
A \(x^2-11x+30=0\)
B \(x^2+11x+30=0\)
C \(x^2-5x+6=0\)
D \(x^2-30x+11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+30=0\)
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=6\) हैं। नई जड़ें (5) और (6) हैं, इसलिए समीकरण \(x^2-11x+30=0\) है।
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यदि \(x^2-9x+14=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-3\) और \(\beta-3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-9x+14=0\), which equation has roots \(\alpha-3\) and \(\beta-3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-3x-4=0\)
B \(x^2+3x-4=0\)
C \(x^2-3x+4=0\)
D \(x^2-9x+14=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-4=0\)
Step 1
Concept
The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-4=0\). The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (7) हैं। नई जड़ें (-1) और (4) होंगी, इसलिए समीकरण \(x^2-3x-4=0\) है।
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यदि \(x^2-6x+5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(3\alpha-2\) और \(3\beta-2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-6x+5=0\), which equation has roots \(3\alpha-2\) and \(3\beta-2\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-14x+13=0\)
B \(x^2-18x+45=0\)
C \(x^2-14x+25=0\)
D \(x^2+14x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-14x+13=0\)
Step 1
Concept
The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-14x+13=0\). The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 3
Exam Tip
मूल जड़ें (1) और (5) हैं, इसलिए नई जड़ें (1) और (13) हैं। उनका समीकरण \(x^2-14x+13=0\) है।
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\(\frac{1}{2}\) और \(\frac{3}{4}\) जड़ों वाला द्विघात समीकरण कौन-सा है?
Which quadratic equation has roots \(\frac{1}{2}\) and \(\frac{3}{4}\)?
#quadratic-roots
#forming-equation
#fractional-roots
A \(8x^2-10x+3=0\)
B \(8x^2+10x+3=0\)
C \(4x^2-10x+3=0\)
D \(8x^2-6x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(8x^2-10x+3=0\)
Step 1
Concept
The sum is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 2
Why this answer is correct
The correct answer is A. \(8x^2-10x+3=0\). The sum is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 3
Exam Tip
जड़ों का योग \(\frac{5}{4}\) और गुणनफल \(\frac{3}{8}\) है। समीकरण \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) को (8) से गुणा करें।
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यदि \(x^2-3x-2=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2,\beta^2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-3x-2=0\), which equation has roots \(\alpha^2,\beta^2\)?
#quadratic-roots
#squared-roots
#new-equation
A \(x^2-13x+4=0\)
B \(x^2+13x+4=0\)
C \(x^2-9x+4=0\)
D \(x^2-13x-4=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+4=0\)
Step 1
Concept
Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+4=0\). Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 3
Exam Tip
\(\alpha+\beta=3\) और \(\alpha\beta=-2\) है। इसलिए \(\alpha^2+\beta^2=13\) और \(\alpha^2\beta^2=4\), अतः समीकरण \(x^2-13x+4=0\) है।
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\(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं। \(\alpha+1,\beta+1\) जड़ों वाला समीकरण कौन-सा है?
The roots of \(x^2-5x+6=0\) are \(\alpha,\beta\). Which equation has roots \(\alpha+1,\beta+1\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-7x+12=0\)
B \(x^2-5x+12=0\)
C \(x^2-6x+8=0\)
D \(x^2+7x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+12=0\)
Step 1
Concept
The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+12=0\). The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (3) हैं, इसलिए नई जड़ें (3) और (4) होंगी। उनका समीकरण \(x^2-7x+12=0\) है।
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यदि \(3x^2-10x+3=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha},\frac{1}{\beta}\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(3x^2-10x+3=0\), which equation has roots \(\frac{1}{\alpha},\frac{1}{\beta}\)?
#quadratic-roots
#reciprocal-roots
#new-equation
A \(3x^2-10x+3=0\)
B \(3x^2+10x+3=0\)
C \(x^2-10x+3=0\)
D \(10x^2-3x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-10x+3=0\)
Step 1
Concept
Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{10}{3}\) और \(\alpha\beta=1\) है। व्युत्क्रम जड़ों का योग \(\frac{10}{3}\) और गुणनफल (1) ही रहता है।
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यदि \(x^2-7x+10=0\) के मूलों को उलटकर नया समीकरण बनाया जाए तो नया मोनिक समीकरण कौन सा होगा?
If a new equation is formed by taking reciprocals of the roots of \(x^2-7x+10=0\), which monic equation is obtained?
#roots
#reciprocal_roots
#new_equation
A \(x^2-\frac{7}{10}x+\frac{1}{10}=0\)
B \(x^2-7x+10=0\)
C \(x^2+\frac{7}{10}x+\frac{1}{10}=0\)
D \(x^2-\frac{1}{10}x+\frac{7}{10}=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-\frac{7}{10}x+\frac{1}{10}=0\)
Step 1
Concept
The old sum is (7) and product is (10). The reciprocal roots have sum \(\frac{7}{10}\) and product \(\frac{1}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-\frac{7}{10}x+\frac{1}{10}=0\). The old sum is (7) and product is (10). The reciprocal roots have sum \(\frac{7}{10}\) and product \(\frac{1}{10}\).
Step 3
Exam Tip
पुराने योग (7) और गुणनफल (10) हैं। उलटे मूलों का योग \(\frac{7}{10}\) और गुणनफल \(\frac{1}{10}\) होगा।
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यदि \(x^2-4x+3=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(3\alpha\) और \(3\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-4x+3=0\), which equation has \(3\alpha\) and \(3\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-12x+27=0\)
B \(x^2-4x+27=0\)
C \(x^2-12x+9=0\)
D \(x^2+12x+27=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-12x+27=0\)
Step 1
Concept
The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+27=0\). The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 3
Exam Tip
पुराने योग (4) और गुणनफल (3) हैं। नए योग (12) और गुणनफल (27) होंगे इसलिए \(x^2-12x+27=0\) है।
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यदि \(x^2-5x+6=0\) के मूलों को उलटकर नया समीकरण बनाया जाए तो नया मोनिक समीकरण कौन सा होगा?
If a new equation is formed by taking reciprocals of the roots of \(x^2-5x+6=0\), which monic equation is obtained?
#roots
#reciprocal_roots
#new_equation
A \(x^2-\frac{5}{6}x+\frac{1}{6}=0\)
B \(x^2-5x+6=0\)
C \(x^2+\frac{5}{6}x+\frac{1}{6}=0\)
D \(x^2-\frac{1}{6}x+\frac{5}{6}=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-\frac{5}{6}x+\frac{1}{6}=0\)
Step 1
Concept
The old sum is (5) and product is (6). The reciprocal roots have sum \(\frac{5}{6}\) and product \(\frac{1}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-\frac{5}{6}x+\frac{1}{6}=0\). The old sum is (5) and product is (6). The reciprocal roots have sum \(\frac{5}{6}\) and product \(\frac{1}{6}\).
Step 3
Exam Tip
पुराने योग (5) और गुणनफल (6) हैं। उलटे मूलों का योग \(\frac{5}{6}\) और गुणनफल \(\frac{1}{6}\) होगा।
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यदि \(x^2-3x+2=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(2\alpha\) और \(2\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-3x+2=0\), which equation has \(2\alpha\) and \(2\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-6x+8=0\)
B \(x^2-3x+8=0\)
C \(x^2-6x+4=0\)
D \(x^2+6x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+8=0\)
Step 1
Concept
The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+8=0\). The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 3
Exam Tip
पुराने योग (3) और गुणनफल (2) हैं। नए योग (6) और गुणनफल (8) होंगे इसलिए \(x^2-6x+8=0\) है।
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यदि किसी द्विघात समीकरण में (a=2), (b=-7), (c=3) हैं, तो समीकरण कौन-सा है?
If a quadratic equation has (a=2), (b=-7), (c=3), which equation is it?
#quadratic-equations
#forming-equation
#coefficients
#medium
A \(2x^2+7x+3=0\)
B \(2x^2-7x+3=0\)
C \(3x^2-7x+2=0\)
D \(2x^2-3x+7=0\)
Explanation opens after your attempt
Correct Answer
B. \(2x^2-7x+3=0\)
Step 1
Concept
Substituting in \(ax^2+bx+c=0\) gives \(2x^2-7x+3=0\). Keep the negative sign of (b).
Step 2
Why this answer is correct
The correct answer is B. \(2x^2-7x+3=0\). Substituting in \(ax^2+bx+c=0\) gives \(2x^2-7x+3=0\). Keep the negative sign of (b).
Step 3
Exam Tip
मानक रूप \(ax^2+bx+c=0\) में मान रखने पर \(2x^2-7x+3=0\) मिलता है। (b) का ऋण चिन्ह साथ रखें।
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यदि किसी समीकरण में लोहे के परमाणु बराबर हैं पर ऑक्सीजन परमाणु बराबर नहीं हैं तो समीकरण कैसा है?
If iron atoms are equal but oxygen atoms are not equal in an equation what is the equation?
#science
#class10
#expert
#unbalanced-equation
#atoms
A पूर्ण संतुलित / Fully balanced
B असंतुलित / Unbalanced
C अवक्षेपित / Precipitated
D ऊष्माक्षेपी / Exothermic
Explanation opens after your attempt
Correct Answer
B. असंतुलित / Unbalanced
Step 1
Concept
Every element must have equal atoms in a balanced equation.
Step 2
Why this answer is correct
Imbalance of even one element is enough.
Step 3
Exam Tip
Therefore the equation is unbalanced when oxygen is unequal. चरण 1: संतुलित समीकरण में हर तत्व की संख्या बराबर होनी चाहिए। चरण 2: केवल एक तत्व का असंतुलन भी पर्याप्त है। चरण 3: इसलिए ऑक्सीजन असमान होने पर समीकरण असंतुलित है।
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कौन सा कथन कंकाली समीकरण और संतुलित समीकरण के बीच अंतर को सही बताता है?
Which statement correctly shows the difference between a skeletal equation and a balanced equation?
#science
#class10
#hard
#skeletal-equation
#balanced-equation
A कंकाली समीकरण में परमाणु असमान हो सकते हैं पर संतुलित समीकरण में समान होते हैं / Atoms may be unequal in a skeletal equation but equal in a balanced equation
B कंकाली समीकरण में उत्पाद नहीं होते / Skeletal equation has no products
C संतुलित समीकरण में अभिकारक नहीं होते / Balanced equation has no reactants
D दोनों हमेशा समान होते हैं / Both are always same
Explanation opens after your attempt
Correct Answer
A. कंकाली समीकरण में परमाणु असमान हो सकते हैं पर संतुलित समीकरण में समान होते हैं / Atoms may be unequal in a skeletal equation but equal in a balanced equation
Step 1
Concept
A skeletal equation gives a framework of formulae.
Step 2
Why this answer is correct
Atom numbers may not be equal in it.
Step 3
Exam Tip
In a balanced equation all elements are equal on both sides. चरण 1: कंकाली समीकरण सूत्रों का ढांचा देता है। चरण 2: इसमें परमाणुओं की संख्या बराबर न हो सकती है। चरण 3: संतुलित समीकरण में सभी तत्व दोनों ओर बराबर होते हैं।
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कंकाली समीकरण को संतुलित समीकरण में बदलने के लिए क्या करना चाहिए?
What should be done to convert a skeletal equation into a balanced equation?
#science
#class10
#medium
#skeletal-equation
#coefficients
A गुणांक लगाकर परमाणु बराबर करने चाहिए / Coefficients should be used to make atoms equal
B रासायनिक सूत्र बदलने चाहिए / Chemical formulae should be changed
C अभिकारक हटाने चाहिए / Reactants should be removed
D तीर उल्टा करना चाहिए / The arrow should be reversed
Explanation opens after your attempt
Correct Answer
A. गुणांक लगाकर परमाणु बराबर करने चाहिए / Coefficients should be used to make atoms equal
Step 1
Concept
A skeletal equation may not have equal atoms.
Step 2
Why this answer is correct
Coefficients are used to make atoms equal on both sides.
Step 3
Exam Tip
This is the correct method of balancing. चरण 1: कंकाली समीकरण में परमाणु बराबर नहीं हो सकते। चरण 2: गुणांक लगाकर दोनों ओर परमाणु बराबर किए जाते हैं। चरण 3: यही संतुलन की सही विधि है।
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कंकाली समीकरण को संतुलित समीकरण में बदलने के लिए क्या किया जाता है?
What is done to convert a skeletal equation into a balanced equation?
#science
#class10
#medium
#skeletal-equation
#coefficients
A रासायनिक सूत्र बदले जाते हैं / Chemical formulae are changed
B गुणांक लगाकर परमाणु बराबर किए जाते हैं / Coefficients are used to make atoms equal
C उत्पाद हटाए जाते हैं / Products are removed
D अभिकारक दाईं ओर लिखे जाते हैं / Reactants are written on the right side
Explanation opens after your attempt
Correct Answer
B. गुणांक लगाकर परमाणु बराबर किए जाते हैं / Coefficients are used to make atoms equal
Step 1
Concept
Atoms may not be equal in a skeletal equation.
Step 2
Why this answer is correct
Coefficients are used to make atoms of each element equal.
Step 3
Exam Tip
This is the correct way to make a balanced equation. चरण 1: कंकाली समीकरण में परमाणु बराबर नहीं हो सकते। चरण 2: गुणांक लगाकर प्रत्येक तत्व के परमाणु बराबर किए जाते हैं। चरण 3: यही संतुलित समीकरण बनाने का सही तरीका है।
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शब्द समीकरण को रासायनिक समीकरण में बदलने के बाद अगला प्रमुख कार्य क्या है?
After converting a word equation into a chemical equation what is the next main task?
#science
#class10
#medium
#word-equation
#balancing
A समीकरण को संतुलित करना / Balancing the equation
B सभी उत्पाद हटाना / Removing all products
C सभी अभिकारक मिटाना / Erasing all reactants
D तीर उल्टा करना / Reversing the arrow
Explanation opens after your attempt
Correct Answer
A. समीकरण को संतुलित करना / Balancing the equation
Step 1
Concept
Correct chemical formulae are first written from the word equation.
Step 2
Why this answer is correct
After writing formulae atoms are counted.
Step 3
Exam Tip
Then coefficients are used to balance the equation. चरण 1: शब्द समीकरण से पहले सही रासायनिक सूत्र लिखे जाते हैं। चरण 2: सूत्र लिखने के बाद परमाणुओं की गिनती की जाती है। चरण 3: फिर गुणांक लगाकर समीकरण संतुलित किया जाता है।
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शब्द समीकरण से रासायनिक समीकरण बनाते समय सबसे बड़ी सावधानी क्या है?
What is the most important care while converting a word equation into a chemical equation?
#science
#class10
#word-equation
#chemical-formulae
A सभी पदार्थों के सही रासायनिक सूत्र लिखना / Writing correct chemical formulae of all substances
B सभी पदार्थों का रंग बदलना / Changing colour of all substances
C तीर को मिटा देना / Removing the arrow
D अभिकारकों को दाईं ओर लिखना / Writing reactants on the right side
Explanation opens after your attempt
Correct Answer
A. सभी पदार्थों के सही रासायनिक सूत्र लिखना / Writing correct chemical formulae of all substances
Step 1
Concept
A word equation is written in names.
Step 2
Why this answer is correct
Correct formulae are necessary in a chemical equation.
Step 3
Exam Tip
A wrong formula can make the whole equation wrong. चरण 1: शब्द समीकरण नामों में लिखा होता है। चरण 2: रासायनिक समीकरण में सही सूत्र लिखना आवश्यक है। चरण 3: गलत सूत्र से पूरा समीकरण गलत हो सकता है।
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शब्द समीकरण को रासायनिक समीकरण में बदलने के लिए क्या लिखना जरूरी है?
What must be written to convert a word equation into a chemical equation?
#science
#class10
#word-equation
#chemical-formulae
A सही रासायनिक सूत्र / Correct chemical formulae
B पदार्थों की कीमत / Price of substances
C पात्रों के नाम / Names of vessels
D प्रयोगशाला का पता / Address of laboratory
Explanation opens after your attempt
Correct Answer
A. सही रासायनिक सूत्र / Correct chemical formulae
Step 1
Concept
A word equation gives names of substances.
Step 2
Why this answer is correct
A chemical equation writes their correct formulae.
Step 3
Exam Tip
A wrong formula can make the whole equation incorrect. चरण 1: शब्द समीकरण पदार्थों के नाम बताता है। चरण 2: रासायनिक समीकरण में उनके सही सूत्र लिखे जाते हैं। चरण 3: गलत सूत्र होने पर पूरा समीकरण गलत हो सकता है।
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शब्द समीकरण से रासायनिक समीकरण बनाते समय पहला कार्य क्या होता है?
What is the first task while converting a word equation into a chemical equation?
#science
#class10
#word-equation
#chemical-formula
A पदार्थों के सही रासायनिक सूत्र लिखना / Writing correct chemical formulae of substances
B रंग बदलना / Changing colours
C पदार्थों को गर्म करना / Heating substances
D तीर हटाना / Removing arrow
Explanation opens after your attempt
Correct Answer
A. पदार्थों के सही रासायनिक सूत्र लिखना / Writing correct chemical formulae of substances
Step 1
Concept
A word equation is written in names.
Step 2
Why this answer is correct
It must be converted into formulae.
Step 3
Exam Tip
A correct equation cannot be made without correct formulae. चरण 1: शब्द समीकरण नामों में लिखा होता है। चरण 2: इसे सूत्रों में बदलना आवश्यक है। चरण 3: सही सूत्र लिखे बिना सही समीकरण नहीं बनता।
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सामान्य द्विघात समीकरण \(ax^2+bx+c=0\) में जड़ें समान हों, तो सही संबंध कौन-सा है?
For the general quadratic equation \(ax^2+bx+c=0\), which relation is correct when the roots are equal?
#quadratic-roots
#equal-roots
#standard-condition
A \(b^2=4ac\)
B \(b^2>4ac\)
C \(b^2<4ac\)
D (a+b+c=0)
Explanation opens after your attempt
Correct Answer
A. \(b^2=4ac\)
Step 1
Concept
For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 2
Why this answer is correct
The correct answer is A. \(b^2=4ac\). For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 3
Exam Tip
समान वास्तविक जड़ों के लिए विविक्तकर \(D=b^2-4ac=0\) होता है। इसलिए \(b^2=4ac\) सही संबंध है।
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यदि \(x^2-6x-16=0\) के मूलों को (1) बढ़ा दिया जाए तो नए मूलों से बना मोनिक समीकरण कौन सा होगा?
If each root of \(x^2-6x-16=0\) is increased by (1), which monic equation is formed from the new roots?
#roots
#transformed_roots
#new_equation
A \(x^2-8x-9=0\)
B \(x^2-6x-16=0\)
C \(x^2-4x-9=0\)
D \(x^2+8x-9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x-9=0\)
Step 1
Concept
The old roots are (8) and (-2). The new roots are (9) and (-1), so the equation is \(x^2-8x-9=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x-9=0\). The old roots are (8) and (-2). The new roots are (9) and (-1), so the equation is \(x^2-8x-9=0\).
Step 3
Exam Tip
पुराने मूल (8) और (-2) हैं। नए मूल (9) और (-1) होंगे इसलिए समीकरण \(x^2-8x-9=0\) है।
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यदि \(x^2-4x-12=0\) के मूलों को (1) बढ़ा दिया जाए तो नए मूलों से बना मोनिक समीकरण कौन सा होगा?
If each root of \(x^2-4x-12=0\) is increased by (1), which monic equation is formed from the new roots?
#roots
#transformed_roots
#new_equation
A \(x^2-6x-15=0\)
B \(x^2-4x-12=0\)
C \(x^2-2x-15=0\)
D \(x^2+6x-15=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x-15=0\)
Step 1
Concept
The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is \(x^2-6x-7=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x-15=0\). The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is \(x^2-6x-7=0\).
Step 3
Exam Tip
पुराने मूल (6) और (-2) हैं। नए मूल (7) और (-1) होंगे इसलिए समीकरण \(x^2-6x-7=0\) नहीं बल्कि \(x^2-6x-7=0\) होता है।
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यदि \(\alpha+\beta=-7\) और \(\alpha\beta=-18\) है तो \(\alpha\) और \(\beta\) के लिए मोनिक समीकरण कौन सा है?
If \(\alpha+\beta=-7\) and \(\alpha\beta=-18\), which monic equation has roots \(\alpha\) and \(\beta\)?
#roots
#equation_from_roots
#sum_product
A \(x^2+7x-18=0\)
B \(x^2-7x-18=0\)
C \(x^2+18x-7=0\)
D \(x^2-18x+7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x-18=0\)
Step 1
Concept
The monic equation is (x-2 -\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+7x-18=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x-18=0\). The monic equation is (x-2 -\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+7x-18=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2 -\(\alpha+\beta\)x+\alpha\beta=0) होता है। इसलिए \(x^2+7x-18=0\) सही है।
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यदि किसी द्विघात समीकरण के मूल (c) और (-c) हैं तो अचर पद और अग्र गुणांक के अनुपात \(\frac{d}{a}\) का मान क्या होगा?
If the roots of a quadratic equation are (c) and (-c), what is the value of the ratio \(\frac{d}{a}\) of constant term to leading coefficient?
#roots
#opposite_roots
#product
A \(-c^2\)
B \(c^2\)
C (0)
D (2c)
Explanation opens after your attempt
Correct Answer
A. \(-c^2\)
Step 1
Concept
\(\frac{d}{a}\) is the product of roots. Here (c(-c)=-c-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(-c^2\). \(\frac{d}{a}\) is the product of roots. Here (c(-c)=-c-2 ).
Step 3
Exam Tip
\(\frac{d}{a}\) मूलों का गुणनफल होता है। यहां (c(-c)=-c-2 ) है।
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यदि द्विघात समीकरण के मूल (6) और \(\frac{1}{6}\) हैं तो उनके बारे में सही कथन कौन सा है?
If the roots of a quadratic equation are (6) and \(\frac{1}{6}\), which statement is correct about them?
#roots
#reciprocal_roots
#concept
A वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
B वे बराबर मूल हैं / They are equal roots
C उनका योग (1) है / Their sum is (1)
D उनका गुणनफल (0) है / Their product is (0)
Explanation opens after your attempt
Correct Answer
A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
Step 1
Concept
\(6\cdot\frac{1}{6}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 2
Why this answer is correct
The correct answer is A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other. \(6\cdot\frac{1}{6}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 3
Exam Tip
\(6\cdot\frac{1}{6}=1\) है इसलिए दोनों व्युत्क्रम मूल हैं। व्युत्क्रम मूलों का गुणनफल (1) होता है।
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यदि (x=3) और (x=8) किसी मोनिक द्विघात समीकरण के मूल हैं तो (x) का गुणांक क्या होगा?
If (x=3) and (x=8) are roots of a monic quadratic equation, what will be the coefficient of (x)?
#roots
#coefficient_from_roots
#monic
A (-11)
B (11)
C (24)
D (-24)
Explanation opens after your attempt
Step 1
Concept
\(The sum of roots is (3+8=11). In a monic equation the coefficient of (x) is (-(\)sum\()=-11).\)
Step 2
Why this answer is correct
\(The correct answer is A. (-11). The sum of roots is (3+8=11). In a monic equation the coefficient of (x) is (-(\)sum\()=-11).\)
Step 3
Exam Tip
मूलों का योग (3+8=11) है। मोनिक समीकरण में (x) का गुणांक (-(योग\()=-11) होता है\)।
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यदि मूलों का योग (0) और गुणनफल (-36) है तो मोनिक समीकरण कौन सा होगा?
If the sum of roots is (0) and product is (-36), which monic equation is formed?
#roots
#equation_from_sum_product
#zero_sum
A \(x^2-36=0\)
B \(x^2+36=0\)
C \(x^2-36x=0\)
D \(x^2+36x=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-36=0\)
Step 1
Concept
The monic equation is (x-2 -(0)x+(-36)=0). Therefore \(x^2-36=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-36=0\). The monic equation is (x-2 -(0)x+(-36)=0). Therefore \(x^2-36=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2 -(0)x+(-36)=0) होगा। इसलिए \(x^2-36=0\) सही है।
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यदि किसी द्विघात समीकरण के मूलों का योग (0) है तो मूलों के बारे में सही कथन कौन सा हो सकता है?
If the sum of roots of a quadratic equation is (0), which statement about the roots can be correct?
#roots
#zero_sum
#reasoning
A मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
B दोनों मूल हमेशा (1) हैं / Both roots are always (1)
C दोनों मूल हमेशा धनात्मक हैं / Both roots are always positive
D मूलों का गुणनफल हमेशा (0) है / The product is always (0)
Explanation opens after your attempt
Correct Answer
A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
Step 1
Concept
If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 2
Why this answer is correct
The correct answer is A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other. If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 3
Exam Tip
यदि \(\alpha+\beta=0\) है तो \(\beta=-\alpha\) होता है। इसलिए मूल विपरीत हो सकते हैं।
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यदि किसी द्विघात समीकरण में (a=1), (b=-20), (c=100) है तो मूल कौन से होंगे?
If a quadratic equation has (a=1), (b=-20), and (c=100), what will be the roots?
#roots
#coefficients
#repeated_root
A (10) और (10) / (10) and (10)
B (-10) और (-10) / (-10) and (-10)
C (20) और (100) / (20) and (100)
D (0) और (10) / (0) and (10)
Explanation opens after your attempt
Correct Answer
A. (10) और (10) / (10) and (10)
Step 1
Concept
The equation is \(x^2-20x+100=0\), which becomes ((x-10)2 =0). Therefore both roots are (10).
Step 2
Why this answer is correct
The correct answer is A. (10) और (10) / (10) and (10). The equation is \(x^2-20x+100=0\), which becomes ((x-10)2 =0). Therefore both roots are (10).
Step 3
Exam Tip
समीकरण \(x^2-20x+100=0\) है जो ((x-10)2 =0) बनता है। इसलिए दोनों मूल (10) हैं।
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यदि (x+6) और (x-2) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x+6) and (x-2) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (-6) और (2) / (-6) and (2)
B (6) और (-2) / (6) and (-2)
C (6) और (2) / (6) and (2)
D (-6) और (-2) / (-6) and (-2)
Explanation opens after your attempt
Correct Answer
A. (-6) और (2) / (-6) and (2)
Step 1
Concept
From (x+6=0), (x=-6), and from (x-2=0), (x=2). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (-6) और (2) / (-6) and (2). From (x+6=0), (x=-6), and from (x-2=0), (x=2). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x+6=0) से (x=-6) और (x-2=0) से (x=2) मिलता है। गुणनखंड से मूल निकालते समय चिन्ह बदलता है।
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यदि द्विघात समीकरण के मूल (4) और (-9) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?
If the roots of a quadratic equation are (4) and (-9), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#roots
#reciprocal_sum
#identity
A \(\frac{5}{36}\)
B -\(\frac{5}{36}\)
C \(\frac{13}{36}\)
D -\(\frac{13}{36}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{36}\)
Step 1
Concept
Here \(\alpha+\beta=-5\) and \(\alpha\beta=-36\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{36}\). Here \(\alpha+\beta=-5\) and \(\alpha\beta=-36\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\).
Step 3
Exam Tip
यहां \(\alpha+\beta=-5\) और \(\alpha\beta=-36\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\) है।
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यदि मूलों का योग (11) और गुणनफल (28) है तो मोनिक द्विघात समीकरण कौन सा होगा?
If the sum of roots is (11) and product is (28), which monic quadratic equation is formed?
#roots
#equation_from_sum_product
#monic
A \(x^2-11x+28=0\)
B \(x^2+11x+28=0\)
C \(x^2-28x+11=0\)
D \(x^2+28x+11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+28=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-11x+28=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-11x+28=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-11x+28=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-11x+28=0) सही है\)।
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यदि किसी द्विघात समीकरण के लिए (D=81) है तो वास्तविक मूलों की प्रकृति क्या होगी?
If (D=81) for a quadratic equation, what will be the nature of real roots?
#roots
#discriminant
#nature
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D एक मूल (0) / One root (0)
Explanation opens after your attempt
Correct Answer
A. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
Since (D=81>0), there will be two distinct real roots. (D>0) indicates different roots.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक मूल / Two distinct real roots. Since (D=81>0), there will be two distinct real roots. (D>0) indicates different roots.
Step 3
Exam Tip
क्योंकि (D=81>0) है इसलिए दो भिन्न वास्तविक मूल होंगे। (D>0) मूलों के अलग होने का संकेत है।
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किस समीकरण के मूल (3) और (-8) हैं?
Which equation has roots (3) and (-8)?
#roots
#equation_from_roots
#mixed_signs
A \(x^2+5x-24=0\)
B \(x^2-5x-24=0\)
C \(x^2+11x+24=0\)
D \(x^2-11x+24=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x-24=0\)
Step 1
Concept
With roots (3) and (-8), we get ((x-3)(x+8)=0). Expanding gives \(x^2+5x-24=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x-24=0\). With roots (3) and (-8), we get ((x-3)(x+8)=0). Expanding gives \(x^2+5x-24=0\).
Step 3
Exam Tip
मूल (3) और (-8) होने पर ((x-3)(x+8)=0) होगा। खोलने पर \(x^2+5x-24=0\) मिलता है।
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यदि किसी द्विघात समीकरण (q(x)=0) में (q(t)=0) है तो (t) क्या कहलाता है?
If (q(t)=0) in a quadratic equation (q(x)=0), what is (t) called?
#roots
#definition
#concept
A मूल / Root
B घात / Degree
C अचर पद / Constant term
D गुणांक / Coefficient
Explanation opens after your attempt
Correct Answer
A. मूल / Root
Step 1
Concept
(q(t)=0) means the equation is true when (x=t). Therefore (t) is a root of the equation.
Step 2
Why this answer is correct
The correct answer is A. मूल / Root. (q(t)=0) means the equation is true when (x=t). Therefore (t) is a root of the equation.
Step 3
Exam Tip
(q(t)=0) का अर्थ है कि (x=t) रखने पर समीकरण सत्य है। इसलिए (t) उस समीकरण का मूल है।
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यदि \(\alpha+\beta=-5\) और \(\alpha\beta=-14\) है तो \(\alpha\) और \(\beta\) के लिए मोनिक समीकरण कौन सा है?
If \(\alpha+\beta=-5\) and \(\alpha\beta=-14\), which monic equation has roots \(\alpha\) and \(\beta\)?
#roots
#equation_from_roots
#sum_product
A \(x^2+5x-14=0\)
B \(x^2-5x-14=0\)
C \(x^2+14x-5=0\)
D \(x^2-14x+5=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x-14=0\)
Step 1
Concept
The monic equation is (x-2 -\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+5x-14=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x-14=0\). The monic equation is (x-2 -\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+5x-14=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2 -\(\alpha+\beta\)x+\alpha\beta=0) होता है। इसलिए \(x^2+5x-14=0\) सही है।
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यदि किसी द्विघात समीकरण के मूल (b) और (-b) हैं तो अचर पद और अग्र गुणांक के अनुपात \(\frac{c}{a}\) का मान क्या होगा?
If the roots of a quadratic equation are (b) and (-b), what is the value of the ratio \(\frac{c}{a}\) of constant term to leading coefficient?
#roots
#opposite_roots
#product
A \(-b^2\)
B \(b^2\)
C (0)
D (2b)
Explanation opens after your attempt
Correct Answer
A. \(-b^2\)
Step 1
Concept
\(\frac{c}{a}\) is the product of roots. Here (b(-b)=-b-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(-b^2\). \(\frac{c}{a}\) is the product of roots. Here (b(-b)=-b-2 ).
Step 3
Exam Tip
\(\frac{c}{a}\) मूलों का गुणनफल होता है। यहां (b(-b)=-b-2 ) है।
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यदि द्विघात समीकरण के मूल (5) और \(\frac{1}{5}\) हैं तो उनके बारे में सही कथन कौन सा है?
If the roots of a quadratic equation are (5) and \(\frac{1}{5}\), which statement is correct about them?
#roots
#reciprocal_roots
#concept
A वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
B वे बराबर मूल हैं / They are equal roots
C उनका योग (1) है / Their sum is (1)
D उनका गुणनफल (0) है / Their product is (0)
Explanation opens after your attempt
Correct Answer
A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
Step 1
Concept
\(5\cdot\frac{1}{5}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 2
Why this answer is correct
The correct answer is A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other. \(5\cdot\frac{1}{5}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 3
Exam Tip
\(5\cdot\frac{1}{5}=1\) है इसलिए दोनों व्युत्क्रम मूल हैं। व्युत्क्रम मूलों का गुणनफल (1) होता है।
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यदि (x=2) और (x=7) किसी मोनिक द्विघात समीकरण के मूल हैं तो (x) का गुणांक क्या होगा?
If (x=2) and (x=7) are roots of a monic quadratic equation, what will be the coefficient of (x)?
#roots
#coefficient_from_roots
#monic
A (-9)
B (9)
C (14)
D (-14)
Explanation opens after your attempt
Step 1
Concept
\(The sum of roots is (2+7=9). In a monic equation the coefficient of (x) is (-(\)sum\()=-9).\)
Step 2
Why this answer is correct
\(The correct answer is A. (-9). The sum of roots is (2+7=9). In a monic equation the coefficient of (x) is (-(\)sum\()=-9).\)
Step 3
Exam Tip
मूलों का योग (2+7=9) है। मोनिक समीकरण में (x) का गुणांक (-(योग\()=-9) होता है\)।
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यदि मूलों का योग (0) और गुणनफल (-25) है तो मोनिक समीकरण कौन सा होगा?
If the sum of roots is (0) and product is (-25), which monic equation is formed?
#roots
#equation_from_sum_product
#zero_sum
A \(x^2-25=0\)
B \(x^2+25=0\)
C \(x^2-25x=0\)
D \(x^2+25x=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-25=0\)
Step 1
Concept
The monic equation is (x-2 -(0)x+(-25)=0). Therefore \(x^2-25=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-25=0\). The monic equation is (x-2 -(0)x+(-25)=0). Therefore \(x^2-25=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2 -(0)x+(-25)=0) होगा। इसलिए \(x^2-25=0\) सही है।
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यदि किसी द्विघात समीकरण के मूलों का गुणनफल ऋणात्मक है तो सही कथन कौन सा है?
If the product of roots of a quadratic equation is negative, which statement is correct?
#roots
#sign_of_roots
#reasoning
A एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative
B दोनों मूल धनात्मक हैं / Both roots are positive
C दोनों मूल ऋणात्मक हैं / Both roots are negative
D दोनों मूल बराबर हैं / Both roots are equal
Explanation opens after your attempt
Correct Answer
A. एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative
Step 1
Concept
A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.
Step 2
Why this answer is correct
The correct answer is A. एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative. A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.
Step 3
Exam Tip
ऋणात्मक गुणनफल तभी होता है जब मूलों के चिन्ह विपरीत हों। इसलिए एक मूल धनात्मक और दूसरा ऋणात्मक होगा।
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यदि किसी द्विघात समीकरण में (a=1), (b=-16), (c=64) है तो मूल कौन से होंगे?
If a quadratic equation has (a=1), (b=-16), and (c=64), what will be the roots?
#roots
#coefficients
#repeated_root
A (8) और (8) / (8) and (8)
B (-8) और (-8) / (-8) and (-8)
C (16) और (64) / (16) and (64)
D (0) और (8) / (0) and (8)
Explanation opens after your attempt
Correct Answer
A. (8) और (8) / (8) and (8)
Step 1
Concept
The equation is \(x^2-16x+64=0\), which becomes ((x-8)2 =0). Therefore both roots are (8).
Step 2
Why this answer is correct
The correct answer is A. (8) और (8) / (8) and (8). The equation is \(x^2-16x+64=0\), which becomes ((x-8)2 =0). Therefore both roots are (8).
Step 3
Exam Tip
समीकरण \(x^2-16x+64=0\) है जो ((x-8)2 =0) बनता है। इसलिए दोनों मूल (8) हैं।
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यदि (x-4) और (x+7) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x-4) and (x+7) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (4) और (-7) / (4) and (-7)
B (-4) और (7) / (-4) and (7)
C (4) और (7) / (4) and (7)
D (-4) और (-7) / (-4) and (-7)
Explanation opens after your attempt
Correct Answer
A. (4) और (-7) / (4) and (-7)
Step 1
Concept
From (x-4=0), (x=4), and from (x+7=0), (x=-7). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (4) और (-7) / (4) and (-7). From (x-4=0), (x=4), and from (x+7=0), (x=-7). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x-4=0) से (x=4) और (x+7=0) से (x=-7) मिलता है। गुणनखंड का चिन्ह उलटकर मूल मिलता है।
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यदि द्विघात समीकरण के मूल (2) और (-7) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?
If the roots of a quadratic equation are (2) and (-7), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#roots
#reciprocal_sum
#identity
A \(\frac{5}{14}\)
B -\(\frac{5}{14}\)
C \(\frac{9}{14}\)
D -\(\frac{9}{14}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{14}\)
Step 1
Concept
Here \(\alpha+\beta=-5\) and \(\alpha\beta=-14\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{14}\). Here \(\alpha+\beta=-5\) and \(\alpha\beta=-14\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\).
Step 3
Exam Tip
यहां \(\alpha+\beta=-5\) और \(\alpha\beta=-14\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\) है।
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यदि मूलों का योग (9) और गुणनफल (18) है तो मोनिक द्विघात समीकरण कौन सा होगा?
If the sum of roots is (9) and product is (18), which monic quadratic equation is formed?
#roots
#equation_from_sum_product
#monic
A \(x^2-9x+18=0\)
B \(x^2+9x+18=0\)
C \(x^2-18x+9=0\)
D \(x^2+18x+9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-9x+18=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-9x+18=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-9x+18=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-9x+18=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-9x+18=0) सही है\)।
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यदि किसी द्विघात समीकरण के लिए (D=64) है तो उसके वास्तविक मूलों की प्रकृति क्या होगी?
If (D=64) for a quadratic equation, what will be the nature of its real roots?
#roots
#discriminant
#nature
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D एक ही मूल (0) / Only one root (0)
Explanation opens after your attempt
Correct Answer
A. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
Since (D=64>0), there will be two distinct real roots. When (D>0), the roots are different.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक मूल / Two distinct real roots. Since (D=64>0), there will be two distinct real roots. When (D>0), the roots are different.
Step 3
Exam Tip
क्योंकि (D=64>0) है इसलिए दो भिन्न वास्तविक मूल होंगे। (D>0) होने पर मूल अलग अलग होते हैं।
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किस समीकरण के मूल (-3) और (6) हैं?
Which equation has roots (-3) and (6)?
#roots
#equation_from_roots
#mixed_signs
A \(x^2-3x-18=0\)
B \(x^2+3x-18=0\)
C \(x^2-9x+18=0\)
D \(x^2+9x+18=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-18=0\)
Step 1
Concept
With roots (-3) and (6), we get ((x+3)(x-6)=0). Expanding it gives \(x^2-3x-18=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-18=0\). With roots (-3) and (6), we get ((x+3)(x-6)=0). Expanding it gives \(x^2-3x-18=0\).
Step 3
Exam Tip
मूल (-3) और (6) होने पर ((x+3)(x-6)=0) होगा। इसे खोलने पर \(x^2-3x-18=0\) मिलता है।
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यदि किसी द्विघात समीकरण (p(x)=0) में (x=a) रखने पर (p(a)=0) हो जाता है तो (a) क्या कहलाता है?
If substituting (x=a) in a quadratic equation (p(x)=0) gives (p(a)=0), what is (a) called?
#roots
#definition
#substitution
A मूल / Root
B अचर पद / Constant term
C मध्य पद / Middle term
D अग्र गुणांक / Leading coefficient
Explanation opens after your attempt
Correct Answer
A. मूल / Root
Step 1
Concept
Since the equation becomes true after substituting (x=a), (a) is a root. In exams check a root by direct substitution.
Step 2
Why this answer is correct
The correct answer is A. मूल / Root. Since the equation becomes true after substituting (x=a), (a) is a root. In exams check a root by direct substitution.
Step 3
Exam Tip
क्योंकि (x=a) रखने पर समीकरण सत्य हो जाता है इसलिए (a) मूल है। परीक्षा में मूल की जांच सीधे प्रतिस्थापन से करें।
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यदि (D<0) है तो किसी द्विघात समीकरण के वास्तविक मूलों की संख्या कितनी होगी?
If (D<0), how many real roots will a quadratic equation have?
#roots
#discriminant
#no_real_roots
A (0)
B (1)
C (2)
D अनगिनत / Infinitely many
Explanation opens after your attempt
Step 1
Concept
When (D<0), there are no real roots. Therefore the number of real roots is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). When (D<0), there are no real roots. Therefore the number of real roots is (0).
Step 3
Exam Tip
(D<0) होने पर वास्तविक मूल नहीं होते। इसलिए वास्तविक मूलों की संख्या (0) होगी।
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यदि \(\alpha+\beta=-3\) और \(\alpha\beta=-10\) है तो \(\alpha\) और \(\beta\) के लिए मोनिक समीकरण कौन सा है?
If \(\alpha+\beta=-3\) and \(\alpha\beta=-10\), which monic equation has roots \(\alpha\) and \(\beta\)?
#roots
#equation_from_roots
#sum_product
A \(x^2+3x-10=0\)
B \(x^2-3x-10=0\)
C \(x^2+10x-3=0\)
D \(x^2-10x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-10=0\)
Step 1
Concept
The monic equation is (x-2 -\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+3x-10=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-10=0\). The monic equation is (x-2 -\(\alpha+\beta\)x+\alpha\beta=0). Therefore \(x^2+3x-10=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2 -\(\alpha+\beta\)x+\alpha\beta=0) होता है। इसलिए \(x^2+3x-10=0\) सही है।
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यदि किसी द्विघात समीकरण के मूल (a) और (-a) हैं तो अचर पद और अग्र गुणांक के अनुपात \(\frac{c}{a_1}\) का मान क्या होगा?
If the roots of a quadratic equation are (a) and (-a), what is the value of the ratio \(\frac{c}{a_1}\) of constant term to leading coefficient?
#roots
#opposite_roots
#product
A \(-a^2\)
B \(a^2\)
C (0)
D (2a)
Explanation opens after your attempt
Correct Answer
A. \(-a^2\)
Step 1
Concept
\(\frac{c}{a_1}\) is the product of roots. Here (a(-a)=-a-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(-a^2\). \(\frac{c}{a_1}\) is the product of roots. Here (a(-a)=-a-2 ).
Step 3
Exam Tip
\(\frac{c}{a_1}\) मूलों का गुणनफल होता है। यहां (a(-a)=-a-2 ) है।
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यदि द्विघात समीकरण के मूल (4) और \(\frac{1}{4}\) हैं तो उनके बारे में सही कथन कौन सा है?
If the roots of a quadratic equation are (4) and \(\frac{1}{4}\), which statement is correct about them?
#roots
#reciprocal_roots
#concept
A वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
B वे बराबर मूल हैं / They are equal roots
C उनका योग (1) है / Their sum is (1)
D उनका गुणनफल (0) है / Their product is (0)
Explanation opens after your attempt
Correct Answer
A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
Step 1
Concept
\(4\cdot\frac{1}{4}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 2
Why this answer is correct
The correct answer is A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other. \(4\cdot\frac{1}{4}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 3
Exam Tip
\(4\cdot\frac{1}{4}=1\) है इसलिए दोनों व्युत्क्रम मूल हैं। व्युत्क्रम मूलों का गुणनफल (1) होता है।
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यदि (x=1) और (x=4) किसी मोनिक द्विघात समीकरण के मूल हैं तो (x) का गुणांक क्या होगा?
If (x=1) and (x=4) are roots of a monic quadratic equation, what will be the coefficient of (x)?
#roots
#coefficient_from_roots
#monic
A (-5)
B (5)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
\(The sum of roots is (1+4=5). In a monic equation the coefficient of (x) is (-(\)sum\()=-5).\)
Step 2
Why this answer is correct
\(The correct answer is A. (-5). The sum of roots is (1+4=5). In a monic equation the coefficient of (x) is (-(\)sum\()=-5).\)
Step 3
Exam Tip
मूलों का योग (1+4=5) है। मोनिक समीकरण में (x) का गुणांक (-(योग\()=-5) होता है\)।
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यदि मूलों का योग (0) और गुणनफल (-16) है तो मोनिक समीकरण कौन सा होगा?
If the sum of roots is (0) and product is (-16), which monic equation is formed?
#roots
#equation_from_sum_product
#zero_sum
A \(x^2-16=0\)
B \(x^2+16=0\)
C \(x^2-16x=0\)
D \(x^2+16x=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-16=0\)
Step 1
Concept
The monic equation is (x-2 -(0)x+(-16)=0). Therefore \(x^2-16=0\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-16=0\). The monic equation is (x-2 -(0)x+(-16)=0). Therefore \(x^2-16=0\) is correct.
Step 3
Exam Tip
मोनिक समीकरण (x-2 -(0)x+(-16)=0) होगा। इसलिए \(x^2-16=0\) सही है।
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यदि किसी द्विघात समीकरण के मूलों का गुणनफल (0) है तो सही कथन कौन सा है?
If the product of roots of a quadratic equation is (0), which statement is correct?
#roots
#zero_product
#reasoning
A कम से कम एक मूल (0) है / At least one root is (0)
B दोनों मूल जरूर (1) हैं / Both roots must be (1)
C दोनों मूल ऋणात्मक हैं / Both roots are negative
D कोई वास्तविक मूल नहीं है / There is no real root
Explanation opens after your attempt
Correct Answer
A. कम से कम एक मूल (0) है / At least one root is (0)
Step 1
Concept
If \(\alpha\beta=0\), then \(\alpha=0\) or \(\beta=0\). If the product is zero, always check for a zero root.
Step 2
Why this answer is correct
The correct answer is A. कम से कम एक मूल (0) है / At least one root is (0). If \(\alpha\beta=0\), then \(\alpha=0\) or \(\beta=0\). If the product is zero, always check for a zero root.
Step 3
Exam Tip
यदि \(\alpha\beta=0\) है तो \(\alpha=0\) या \(\beta=0\) होगा। गुणनफल शून्य हो तो शून्य मूल जरूर देखें।
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यदि किसी द्विघात समीकरण में (a=1), (b=-10), (c=25) है तो मूल कौन से होंगे?
If a quadratic equation has (a=1), (b=-10), and (c=25), what will be the roots?
#roots
#coefficients
#repeated_root
A (5) और (5) / (5) and (5)
B (-5) और (-5) / (-5) and (-5)
C (10) और (25) / (10) and (25)
D (0) और (5) / (0) and (5)
Explanation opens after your attempt
Correct Answer
A. (5) और (5) / (5) and (5)
Step 1
Concept
The equation is \(x^2-10x+25=0\), which becomes ((x-5)2 =0). Therefore both roots are (5).
Step 2
Why this answer is correct
The correct answer is A. (5) और (5) / (5) and (5). The equation is \(x^2-10x+25=0\), which becomes ((x-5)2 =0). Therefore both roots are (5).
Step 3
Exam Tip
समीकरण \(x^2-10x+25=0\) है जो ((x-5)2 =0) बनता है। इसलिए दोनों मूल (5) हैं।
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यदि (x+2) और (x-5) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x+2) and (x-5) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (-2) और (5) / (-2) and (5)
B (2) और (-5) / (2) and (-5)
C (2) और (5) / (2) and (5)
D (-2) और (-5) / (-2) and (-5)
Explanation opens after your attempt
Correct Answer
A. (-2) और (5) / (-2) and (5)
Step 1
Concept
From (x+2=0), (x=-2), and from (x-5=0), (x=5). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (-2) और (5) / (-2) and (5). From (x+2=0), (x=-2), and from (x-5=0), (x=5). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x+2=0) से (x=-2) और (x-5=0) से (x=5) मिलता है। गुणनखंड का चिन्ह उलटकर मूल मिलता है।
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यदि द्विघात समीकरण के मूल (3) और (-4) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?
If the roots of a quadratic equation are (3) and (-4), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#roots
#reciprocal_sum
#identity
A -\(\frac{1}{12}\)
B \(\frac{1}{12}\)
C -\(\frac{7}{12}\)
D \(\frac{7}{12}\)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{1}{12}\)
Step 1
Concept
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (-1) and product is (-12), so the value is \(\frac{1}{12}\).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{1}{12}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (-1) and product is (-12), so the value is \(\frac{1}{12}\).
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) है। यहां \(\frac{-1}{-12}\) नहीं बल्कि \(-\frac{1}{12}\) क्योंकि योग (-1) और गुणनफल (-12) है।
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यदि मूलों का योग (7) और गुणनफल (10) है तो मोनिक द्विघात समीकरण कौन सा होगा?
If the sum of roots is (7) and product is (10), which monic quadratic equation is formed?
#roots
#equation_from_sum_product
#monic
A \(x^2-7x+10=0\)
B \(x^2+7x+10=0\)
C \(x^2-10x+7=0\)
D \(x^2+10x+7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-7x+10=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-7x+10=0) सही है\)।
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यदि किसी द्विघात समीकरण के लिए (D=49) है तो उसके वास्तविक मूलों की प्रकृति क्या होगी?
If (D=49) for a quadratic equation, what will be the nature of its real roots?
#roots
#discriminant
#nature
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D केवल एक मूल (0) / Only one root (0)
Explanation opens after your attempt
Correct Answer
A. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
Since (D=49>0), two distinct real roots will be obtained. When (D>0), the roots are different.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक मूल / Two distinct real roots. Since (D=49>0), two distinct real roots will be obtained. When (D>0), the roots are different.
Step 3
Exam Tip
क्योंकि (D=49>0) है इसलिए दो भिन्न वास्तविक मूल मिलेंगे। (D>0) होने पर मूल अलग अलग होते हैं।
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यदि किसी द्विघात समीकरण के मूल (r) और (r) हैं तो उनके गुणनफल का मान क्या होगा?
If the roots of a quadratic equation are (r) and (r), what will be their product?
#roots
#equal_roots
#product
A (2r)
B (0)
C \(r^2\)
D (r)
Explanation opens after your attempt
Correct Answer
C. \(r^2\)
Step 1
Concept
The product of (r) and (r) is \(r^2\). For equal roots, remember sum (2r) and product \(r^2\).
Step 2
Why this answer is correct
The correct answer is C. \(r^2\). The product of (r) and (r) is \(r^2\). For equal roots, remember sum (2r) and product \(r^2\).
Step 3
Exam Tip
(r) और (r) का गुणनफल \(r^2\) होता है। बराबर मूलों में योग (2r) और गुणनफल \(r^2\) याद रखें।
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जिस मोनिक द्विघात समीकरण के मूलों का योग (10) और गुणनफल (21) है वह कौन सा है?
Which monic quadratic equation has sum of roots (10) and product of roots (21)?
#roots
#equation_from_sum_product
#monic
A \(x^2+10x+21=0\)
B \(x^2-10x+21=0\)
C \(x^2-21x+10=0\)
D \(x^2+21x+10=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-10x+21=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-10x+21=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-10x+21=0) सही है\)।
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यदि किसी समीकरण के मूल (4) और (-7) हैं तो उनका योग क्या है?
If the roots of an equation are (4) and (-7), what is their sum?
#roots
#sum
#negative_root
A (11)
B (-3)
C (3)
D (-11)
Explanation opens after your attempt
Step 1
Concept
The sum is (4+(-7)=-3). Be careful with the sign while adding a negative number.
Step 2
Why this answer is correct
The correct answer is B. (-3). The sum is (4+(-7)=-3). Be careful with the sign while adding a negative number.
Step 3
Exam Tip
योग (4+(-7)=-3) है। ऋणात्मक संख्या जोड़ते समय चिन्ह का ध्यान रखें।
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किस समीकरण के मूल (6) और (-2) हैं?
Which equation has roots (6) and (-2)?
#roots
#equation_from_roots
#application
A \(x^2-4x-12=0\)
B \(x^2+4x-12=0\)
C \(x^2-8x+12=0\)
D \(x^2+8x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x-12=0\)
Step 1
Concept
With roots (6) and (-2), we get ((x-6)(x+2)=0). Expanding it gives \(x^2-4x-12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x-12=0\). With roots (6) and (-2), we get ((x-6)(x+2)=0). Expanding it gives \(x^2-4x-12=0\).
Step 3
Exam Tip
मूल (6) और (-2) होने पर ((x-6)(x+2)=0) होगा। इसे खोलने पर \(x^2-4x-12=0\) मिलता है।
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किस समीकरण के मूल (-4) और (2) हैं?
Which equation has roots (-4) and (2)?
#roots
#equation_from_roots
#mixed_signs
A \(x^2+2x-8=0\)
B \(x^2-2x-8=0\)
C \(x^2+6x+8=0\)
D \(x^2-6x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x-8=0\)
Step 1
Concept
With roots (-4) and (2), we get ((x+4)(x-2)=0). Expanding it gives \(x^2+2x-8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x-8=0\). With roots (-4) and (2), we get ((x+4)(x-2)=0). Expanding it gives \(x^2+2x-8=0\).
Step 3
Exam Tip
मूल (-4) और (2) होने पर ((x+4)(x-2)=0) होगा। इसे खोलने पर \(x^2+2x-8=0\) मिलता है।
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किस स्थिति में द्विघात समीकरण के दो भिन्न वास्तविक मूल होते हैं?
In which condition does a quadratic equation have two distinct real roots?
#roots
#discriminant
#distinct_roots
A (D=0)
B (D<0)
C (D>0)
D (a=0)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0) is required. This is a direct rule to check the nature.
Step 2
Why this answer is correct
The correct answer is C. (D>0). For two distinct real roots, (D>0) is required. This is a direct rule to check the nature.
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए (D>0) होना चाहिए। यह प्रकृति जांचने का सीधा नियम है।
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यदि (D=0) है तो द्विघात समीकरण के वास्तविक मूल कैसे होते हैं?
If (D=0), how are the real roots of a quadratic equation?
#roots
#discriminant
#equal_roots
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D चार वास्तविक मूल / Four real roots
Explanation opens after your attempt
Correct Answer
B. दो बराबर वास्तविक मूल / Two equal real roots
Step 1
Concept
When (D=0), the two real roots are equal. Check \(D=b^2-4ac\) for the nature of roots.
Step 2
Why this answer is correct
The correct answer is B. दो बराबर वास्तविक मूल / Two equal real roots. When (D=0), the two real roots are equal. Check \(D=b^2-4ac\) for the nature of roots.
Step 3
Exam Tip
(D=0) होने पर दोनों वास्तविक मूल बराबर होते हैं। मूलों की प्रकृति के लिए \(D=b^2-4ac\) देखें।
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किस समीकरण के मूल (0) और (8) हैं?
Which equation has roots (0) and (8)?
#roots
#equation_from_roots
#zero_root
A \(x^2-8x=0\)
B \(x^2+8x=0\)
C \(x^2-64=0\)
D \(x^2+64=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x=0\)
Step 1
Concept
With roots (0) and (8), the equation is (x(x-8)=0). Expanding it gives \(x^2-8x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x=0\). With roots (0) and (8), the equation is (x(x-8)=0). Expanding it gives \(x^2-8x=0\).
Step 3
Exam Tip
मूल (0) और (8) होने पर समीकरण (x(x-8)=0) होगा। इसे खोलने पर \(x^2-8x=0\) मिलता है।
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यदि किसी द्विघात समीकरण के मूल (r) और (-r) हैं तो उनके योग का मान क्या होगा?
If the roots of a quadratic equation are (r) and (-r), what is their sum?
#roots
#opposite_roots
#sum
A (r)
B (2r)
C (0)
D \(r^2\)
Explanation opens after your attempt
Step 1
Concept
(r+(-r)=0). The sum of opposite roots is always (0).
Step 2
Why this answer is correct
The correct answer is C. (0). (r+(-r)=0). The sum of opposite roots is always (0).
Step 3
Exam Tip
(r+(-r)=0) होता है। विपरीत मूलों का योग हमेशा (0) होता है।
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जिस मोनिक द्विघात समीकरण के मूलों का योग (-9) और गुणनफल (20) है वह कौन सा है?
Which monic quadratic equation has sum of roots (-9) and product of roots (20)?
#roots
#equation_from_sum_product
#monic
A \(x^2+9x+20=0\)
B \(x^2-9x+20=0\)
C \(x^2+20x+9=0\)
D \(x^2-20x+9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+9x+20=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+9x+20=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2+9x+20=0) सही है\)।
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यदि किसी समीकरण के मूल (2) और (-5) हैं तो उनका योग क्या है?
If the roots of an equation are (2) and (-5), what is their sum?
#roots
#sum
#negative_root
A (7)
B (-3)
C (3)
D (-7)
Explanation opens after your attempt
Step 1
Concept
The sum is (2+(-5)=-3). Do not forget the sign while adding a negative root.
Step 2
Why this answer is correct
The correct answer is B. (-3). The sum is (2+(-5)=-3). Do not forget the sign while adding a negative root.
Step 3
Exam Tip
योग (2+(-5)=-3) है। ऋणात्मक मूल जोड़ते समय चिन्ह न भूलें।
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किस समीकरण के मूल (5) और (-3) हैं?
Which equation has roots (5) and (-3)?
#roots
#equation_from_roots
#application
A \(x^2-2x-15=0\)
B \(x^2+2x-15=0\)
C \(x^2-8x+15=0\)
D \(x^2+8x+15=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-2x-15=0\)
Step 1
Concept
With roots (5) and (-3), we get ((x-5)(x+3)=0). Expanding it gives \(x^2-2x-15=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-2x-15=0\). With roots (5) and (-3), we get ((x-5)(x+3)=0). Expanding it gives \(x^2-2x-15=0\).
Step 3
Exam Tip
मूल (5) और (-3) होने पर ((x-5)(x+3)=0) होगा। इसे खोलने पर \(x^2-2x-15=0\) मिलता है।
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किस समीकरण के मूल (-2) और (6) हैं?
Which equation has roots (-2) and (6)?
#roots
#equation_from_roots
#mixed_signs
A \(x^2-4x-12=0\)
B \(x^2+4x-12=0\)
C \(x^2-8x+12=0\)
D \(x^2+8x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x-12=0\)
Step 1
Concept
With roots (-2) and (6), we get ((x+2)(x-6)=0). Expanding it gives \(x^2-4x-12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x-12=0\). With roots (-2) and (6), we get ((x+2)(x-6)=0). Expanding it gives \(x^2-4x-12=0\).
Step 3
Exam Tip
मूल (-2) और (6) होने पर ((x+2)(x-6)=0) होगा। इसे खोलने पर \(x^2-4x-12=0\) मिलता है।
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किस स्थिति में द्विघात समीकरण के कोई वास्तविक मूल नहीं होते?
In which condition does a quadratic equation have no real roots?
#roots
#discriminant
#condition
A (D>0)
B (D=0)
C (D<0)
D \(D\ge0\)
Explanation opens after your attempt
Step 1
Concept
When (D<0), there are no real roots. This is a direct rule for the nature of roots.
Step 2
Why this answer is correct
The correct answer is C. (D<0). When (D<0), there are no real roots. This is a direct rule for the nature of roots.
Step 3
Exam Tip
जब (D<0) होता है तब वास्तविक मूल नहीं होते। यह मूलों की प्रकृति का सीधा नियम है।
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यदि (D>0) है तो द्विघात समीकरण के वास्तविक मूल कैसे होते हैं?
If (D>0), how are the real roots of a quadratic equation?
#roots
#discriminant
#nature
A दो बराबर वास्तविक मूल / Two equal real roots
B दो भिन्न वास्तविक मूल / Two distinct real roots
C कोई वास्तविक मूल नहीं / No real root
D केवल एक अचर पद / Only one constant term
Explanation opens after your attempt
Correct Answer
B. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
When (D>0), two distinct real roots are obtained. To know the nature of roots, check \(D=b^2-4ac\).
Step 2
Why this answer is correct
The correct answer is B. दो भिन्न वास्तविक मूल / Two distinct real roots. When (D>0), two distinct real roots are obtained. To know the nature of roots, check \(D=b^2-4ac\).
Step 3
Exam Tip
(D>0) होने पर दो भिन्न वास्तविक मूल मिलते हैं। मूलों की प्रकृति जानने के लिए \(D=b^2-4ac\) देखें।
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किस समीकरण के मूल (0) और (-6) हैं?
Which equation has roots (0) and (-6)?
#roots
#equation_from_roots
#zero_root
A \(x^2+6x=0\)
B \(x^2-6x=0\)
C \(x^2+36=0\)
D \(x^2-36=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+6x=0\)
Step 1
Concept
With roots (0) and (-6), the equation is (x(x+6)=0). Expanding it gives \(x^2+6x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+6x=0\). With roots (0) and (-6), the equation is (x(x+6)=0). Expanding it gives \(x^2+6x=0\).
Step 3
Exam Tip
मूल (0) और (-6) होने पर समीकरण (x(x+6)=0) होगा। इसे खोलने पर \(x^2+6x=0\) मिलता है।
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किसी द्विघात समीकरण (p(x)=0) का मूल वह संख्या है जिससे (p(x)) का मान क्या हो जाता है?
A root of a quadratic equation (p(x)=0) is a number for which the value of (p(x)) becomes what?
#roots
#definition
#concept
A (1)
B (0)
C (-1)
D (p)
Explanation opens after your attempt
Step 1
Concept
When a root is substituted, (p(x)=0) becomes true. In exams always check a root by substitution.
Step 2
Why this answer is correct
The correct answer is B. (0). When a root is substituted, (p(x)=0) becomes true. In exams always check a root by substitution.
Step 3
Exam Tip
मूल रखने पर (p(x)=0) बनता है। परीक्षा में मूल की जांच हमेशा प्रतिस्थापन से करें।
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यदि किसी द्विघात समीकरण के मूल एक दूसरे के व्युत्क्रम हैं तो उनके गुणनफल का मान क्या होगा?
If the roots of a quadratic equation are reciprocals of each other then what is their product?
#roots
#reciprocal_roots
#product
A (0)
B (1)
C (-1)
D (2)
Explanation opens after your attempt
Step 1
Concept
If the roots are (r) and \(\frac{1}{r}\) their product is (1). For reciprocal roots remember the product is (1).
Step 2
Why this answer is correct
The correct answer is B. (1). If the roots are (r) and \(\frac{1}{r}\) their product is (1). For reciprocal roots remember the product is (1).
Step 3
Exam Tip
यदि मूल (r) और \(\frac{1}{r}\) हों तो गुणनफल (1) होता है। व्युत्क्रम मूलों में गुणनफल तुरंत (1) याद रखें।
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जिस द्विघात समीकरण के मूलों का योग (6) और गुणनफल (8) है वह कौन सा है?
Which quadratic equation has sum of roots (6) and product of roots (8)?
#roots
#equation_from_sum_product
#formula
A \(x^2+6x+8=0\)
B \(x^2-6x+8=0\)
C \(x^2-8x+6=0\)
D \(x^2+8x+6=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-6x+8=0\)
Step 1
Concept
\(The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-6x+8=0). The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 3
Exam Tip
\(मानक रूप (x^2-(\)योग)x+गुणनफल=0) है इसलिए \(x^2-6x+8=0\)। योग वाले पद का चिन्ह बदलता है।
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यदि किसी समीकरण के मूल (1) और (-4) हैं तो उनका योग क्या है?
If the roots of an equation are (1) and (-4) then what is their sum?
#roots
#sum
#negative_root
A (5)
B (-3)
C (3)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum is (1+(-4)=-3). Be careful with the sign while adding a negative number.
Step 2
Why this answer is correct
The correct answer is B. (-3). The sum is (1+(-4)=-3). Be careful with the sign while adding a negative number.
Step 3
Exam Tip
योग (1+(-4)=-3) है। ऋणात्मक संख्या जोड़ते समय चिन्ह का ध्यान रखें।
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किस समीकरण के मूल (4) और (-1) हैं?
Which equation has roots (4) and (-1)?
#roots
#equation_from_roots
#mixed_signs
A \(x^2-3x-4=0\)
B \(x^2+3x-4=0\)
C \(x^2-5x+4=0\)
D \(x^2+x-4=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-4=0\)
Step 1
Concept
With roots (4) and (-1) we get ((x-4)(x+1)=0). Expanding it gives \(x^2-3x-4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-4=0\). With roots (4) and (-1) we get ((x-4)(x+1)=0). Expanding it gives \(x^2-3x-4=0\).
Step 3
Exam Tip
मूल (4) और (-1) होने पर ((x-4)(x+1)=0) मिलता है। इसे खोलने पर \(x^2-3x-4=0\) है।
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यदि किसी द्विघात समीकरण के वास्तविक मूलों का गुणनफल ऋणात्मक है तो मूलों के चिन्ह कैसे होंगे?
If the product of real roots of a quadratic equation is negative then how are the signs of the roots?
#roots
#sign_of_roots
#reasoning
A दोनों धनात्मक / Both positive
B दोनों ऋणात्मक / Both negative
C एक धनात्मक और एक ऋणात्मक / One positive and one negative
D दोनों शून्य / Both zero
Explanation opens after your attempt
Correct Answer
C. एक धनात्मक और एक ऋणात्मक / One positive and one negative
Step 1
Concept
A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 2
Why this answer is correct
The correct answer is C. एक धनात्मक और एक ऋणात्मक / One positive and one negative. A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 3
Exam Tip
ऋणात्मक गुणनफल तभी मिलता है जब एक मूल धनात्मक और दूसरा ऋणात्मक हो। \(\alpha\beta<0\) संकेतों की जांच का छोटा संकेत है।
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किस समीकरण के मूल (2) और (5) हैं?
Which equation has roots (2) and (5)?
#roots
#equation_from_roots
#factorisation
A \(x^2-7x+10=0\)
B \(x^2+7x+10=0\)
C \(x^2-3x+10=0\)
D \(x^2+3x-10=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
The equation ((x-2)(x-5)=0) gives \(x^2-7x+10=0\). You can also check the sum and product of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+10=0\). The equation ((x-2)(x-5)=0) gives \(x^2-7x+10=0\). You can also check the sum and product of roots.
Step 3
Exam Tip
समीकरण ((x-2)(x-5)=0) से \(x^2-7x+10=0\) मिलता है। मूलों का योग और गुणनफल भी जांच सकते हैं।
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किस स्थिति में द्विघात समीकरण के वास्तविक मूल होते हैं?
In which condition does a quadratic equation have real roots?
#roots
#real_roots
#discriminant
A (D<0)
B \(D\ge 0\)
C (D=-1)
D (a=0)
Explanation opens after your attempt
Correct Answer
B. \(D\ge 0\)
Step 1
Concept
For real roots \(D\ge 0\) is required. If (D<0) there are no real roots.
Step 2
Why this answer is correct
The correct answer is B. \(D\ge 0\). For real roots \(D\ge 0\) is required. If (D<0) there are no real roots.
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\ge 0\) होना चाहिए। यदि (D<0) हो तो वास्तविक मूल नहीं मिलते।
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यदि किसी द्विघात समीकरण के दो बराबर वास्तविक मूल हों तो डिस्क्रिमिनेंट (D) का मान कैसा होता है?
If a quadratic equation has two equal real roots then what is the value of discriminant (D)?
#roots
#discriminant
#equal_roots
A (D>0)
B (D=0)
C (D<0)
D (D=1)
Explanation opens after your attempt
Step 1
Concept
For equal real roots (D=0). The discriminant quickly tells the nature of roots.
Step 2
Why this answer is correct
The correct answer is B. (D=0). For equal real roots (D=0). The discriminant quickly tells the nature of roots.
Step 3
Exam Tip
बराबर वास्तविक मूलों के लिए (D=0) होता है। डिस्क्रिमिनेंट से मूलों की प्रकृति जल्दी पता चलती है।
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