Question 1/21
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 2/21
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 3/21
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 4/21
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 5/21
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 6/21
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 7/21
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 8/21
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 9/21
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 10/21
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 11/21
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 12/21
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 13/21
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 14/21
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 15/21
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 16/21
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 17/21
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 18/21
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 19/21
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 20/21
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 21/21
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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