22 results found for "equation_from_roots" in Class 10.
Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि (x-2 -(a+2)x+2a=0) के मूल (2) और (a) हैं तो कौन सा कारण सही है?
If roots of (x-2 -(a+2)x+2a=0) are (2) and (a), which reason is correct?
#roots
#reasoning
#equation_from_roots
A योग (a+2) और गुणनफल (2a) है / Sum is (a+2) and product is (2a)
B योग (2a) और गुणनफल (a+2) है / Sum is (2a) and product is (a+2)
C दोनों मूल बराबर हैं / Both roots are equal
D डिस्क्रिमिनेंट हमेशा ऋणात्मक है / Discriminant is always negative
Explanation opens after your attempt
Correct Answer
A. योग (a+2) और गुणनफल (2a) है / Sum is (a+2) and product is (2a)
Step 1
Concept
The roots (2) and (a) have sum (a+2) and product (2a). Therefore the given monic equation is correct.
Step 2
Why this answer is correct
The correct answer is A. योग (a+2) और गुणनफल (2a) है / Sum is (a+2) and product is (2a). The roots (2) and (a) have sum (a+2) and product (2a). Therefore the given monic equation is correct.
Step 3
Exam Tip
मूल (2) और (a) का योग (a+2) तथा गुणनफल (2a) है। इसलिए दिया गया मोनिक समीकरण सही बनता है।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2+px+q=0\) के मूल (-3) और (-4) हैं तो (p+q) का मान क्या है?
If roots of \(x^2+px+q=0\) are (-3) and (-4), what is the value of (p+q)?
#roots
#parameter
#equation_from_roots
A (19)
B (12)
C (7)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (-7), so (p=7), and the product gives (q=12). Hence (p+q=19).
Step 2
Why this answer is correct
The correct answer is A. (19). The sum of roots is (-7), so (p=7), and the product gives (q=12). Hence (p+q=19).
Step 3
Exam Tip
मूलों का योग (-7) है इसलिए (p=7) और गुणनफल (q=12) है। अतः (p+q=19) है।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि (x-2 -(a+1)x+a=0) के मूल (1) और (a) हैं तो यह किस कारण सही है?
If roots of (x-2 -(a+1)x+a=0) are (1) and (a), why is it correct?
#roots
#reasoning
#equation_from_roots
A योग (a+1) और गुणनफल (a) है / Sum is (a+1) and product is (a)
B योग (a) और गुणनफल (a+1) है / Sum is (a) and product is (a+1)
C दोनों मूल बराबर हैं / Both roots are equal
D डिस्क्रिमिनेंट हमेशा ऋणात्मक है / Discriminant is always negative
Explanation opens after your attempt
Correct Answer
A. योग (a+1) और गुणनफल (a) है / Sum is (a+1) and product is (a)
Step 1
Concept
The roots (1) and (a) have sum (a+1) and product (a). Therefore the monic equation is (x-2 -(a+1)x+a=0).
Step 2
Why this answer is correct
The correct answer is A. योग (a+1) और गुणनफल (a) है / Sum is (a+1) and product is (a). The roots (1) and (a) have sum (a+1) and product (a). Therefore the monic equation is (x-2 -(a+1)x+a=0).
Step 3
Exam Tip
मूल (1) और (a) का योग (a+1) तथा गुणनफल (a) है। इसलिए मोनिक समीकरण (x-2 -(a+1)x+a=0) बनता है।
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Question
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2+px+q=0\) के मूल (-2) और (-5) हैं तो (p+q) का मान क्या है?
If roots of \(x^2+px+q=0\) are (-2) and (-5), what is the value of (p+q)?
#roots
#parameter
#equation_from_roots
A (17)
B (3)
C (10)
D (7)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (-7), so (-p=-7) gives (p=7), and (q=10). Hence (p+q=17).
Step 2
Why this answer is correct
The correct answer is A. (17). The sum of roots is (-7), so (-p=-7) gives (p=7), and (q=10). Hence (p+q=17).
Step 3
Exam Tip
मूलों का योग (-7) है इसलिए (-p=-7) से (p=7) और (q=10)। अतः (p+q=17) है।
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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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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Question
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
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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Question
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
किस समीकरण के मूल (0) और (5) हैं?
Which equation has roots (0) and (5)?
#roots
#equation_from_roots
#zero_root
A \(x^2-5x=0\)
B \(x^2+5x=0\)
C \(x^2-25=0\)
D \(x^2+25=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-5x=0\)
Step 1
Concept
With roots (0) and (5) the equation is (x(x-5)=0) that is \(x^2-5x=0\). Remember the form (x-r) while forming an equation from roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-5x=0\). With roots (0) and (5) the equation is (x(x-5)=0) that is \(x^2-5x=0\). Remember the form (x-r) while forming an equation from roots.
Step 3
Exam Tip
मूल (0) और (5) होने पर समीकरण (x(x-5)=0) अर्थात \(x^2-5x=0\) होगा। मूलों से समीकरण बनाते समय (x-r) रूप याद रखें।
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Question
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि मूल (4) और (6) हैं, तो मूलों का गुणनफल क्या है?
If the roots are (4) and (6), what is the product of the roots?
#quadratic-equations
#roots
#product
#medium
A (10)
B (24)
C (2)
D (-24)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(4\cdot6=24\). Sum and product are both useful while forming an equation from roots.
Step 2
Why this answer is correct
The correct answer is B. (24). The product of roots is \(4\cdot6=24\). Sum and product are both useful while forming an equation from roots.
Step 3
Exam Tip
मूलों का गुणनफल \(4\cdot6=24\) है। मूलों से समीकरण बनाते समय योग और गुणनफल दोनों उपयोगी होते हैं।
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