Question 1/25
Medium Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2-4x-32=0\) के लिए सही कथन कौनसा है?
Which statement is correct for \(x^2-4x-32=0\)?
#quadratic
#roots
#mixed-signs
A मूल (8) और (-4) हैं / The roots are (8) and (-4)
B मूल (-8) और (4) हैं / The roots are (-8) and (4)
C मूल (4) और (32) हैं / The roots are (4) and (32)
D मूल (-4) और (-32) हैं / The roots are (-4) and (-32)
Explanation opens after your attempt
Correct Answer
A. मूल (8) और (-4) हैं / The roots are (8) and (-4)
Step 1
Concept
(x-2 -4x-32=(x-8)(x+4)), so the roots are (8) and (-4). In exams, the larger value decides the sign of the middle term.
Step 2
Why this answer is correct
The correct answer is A. मूल (8) और (-4) हैं / The roots are (8) and (-4). (x-2 -4x-32=(x-8)(x+4)), so the roots are (8) and (-4). In exams, the larger value decides the sign of the middle term.
Step 3
Exam Tip
(x-2 -4x-32=(x-8)(x+4)), इसलिए मूल (8) और (-4) हैं। परीक्षा में बड़ा मान मध्य पद का चिन्ह तय करता है।
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Question 2/25
Medium Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(4x^2-19x-5=0\) के मूल क्या हैं?
What are the roots of \(4x^2-19x-5=0\)?
#quadratic
#factorisation
#mixed-signs
A \(x=5,-\frac{1}{4}\)
B \(x=-5,\frac{1}{4}\)
C (x=4,-5)
D \(x=\frac{5}{4},-1\)
Explanation opens after your attempt
Correct Answer
A. \(x=5,-\frac{1}{4}\)
Step 1
Concept
(4x-2 -19x-5=(4x+1)(x-5)), so the roots are (5) and \(-\frac{1}{4}\). In exams, reverse the signs from linear factors to write roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=5,-\frac{1}{4}\). (4x-2 -19x-5=(4x+1)(x-5)), so the roots are (5) and \(-\frac{1}{4}\). In exams, reverse the signs from linear factors to write roots.
Step 3
Exam Tip
(4x-2 -19x-5=(4x+1)(x-5)), इसलिए मूल (5) और \(-\frac{1}{4}\) हैं। परीक्षा में रैखिक गुणनखंडों के चिन्ह उलटकर मूल लिखें।
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Question 3/25
Medium Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
\(x^2-x-30=0\) के लिए सही कथन कौनसा है?
Which statement is correct for \(x^2-x-30=0\)?
#quadratic
#roots
#mixed-signs
A मूल (6) और (-5) हैं / The roots are (6) and (-5)
B मूल (-6) और (5) हैं / The roots are (-6) and (5)
C मूल (1) और (30) हैं / The roots are (1) and (30)
D मूल (-1) और (-30) हैं / The roots are (-1) and (-30)
Explanation opens after your attempt
Correct Answer
A. मूल (6) और (-5) हैं / The roots are (6) and (-5)
Step 1
Concept
(x-2 -x-30=(x-6)(x+5)), so the roots are (6) and (-5). In exams, the larger value decides the sign of the middle term.
Step 2
Why this answer is correct
The correct answer is A. मूल (6) और (-5) हैं / The roots are (6) and (-5). (x-2 -x-30=(x-6)(x+5)), so the roots are (6) and (-5). In exams, the larger value decides the sign of the middle term.
Step 3
Exam Tip
(x-2 -x-30=(x-6)(x+5)), इसलिए मूल (6) और (-5) हैं। परीक्षा में बड़ा मान मध्य पद का चिन्ह तय करता है।
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Question 4/25
Medium Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
\(5x^2-13x-6=0\) के मूल क्या हैं?
What are the roots of \(5x^2-13x-6=0\)?
#quadratic
#factorisation
#mixed-signs
A \(x=3,-\frac{2}{5}\)
B \(x=-3,\frac{2}{5}\)
C (x=5,-6)
D \(x=\frac{3}{5},-2\)
Explanation opens after your attempt
Correct Answer
A. \(x=3,-\frac{2}{5}\)
Step 1
Concept
(5x-2 -13x-6=(5x+2)(x-3)), so the roots are (3) and \(-\frac{2}{5}\). In exams, reverse the signs from linear factors to write roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=3,-\frac{2}{5}\). (5x-2 -13x-6=(5x+2)(x-3)), so the roots are (3) and \(-\frac{2}{5}\). In exams, reverse the signs from linear factors to write roots.
Step 3
Exam Tip
(5x-2 -13x-6=(5x+2)(x-3)), इसलिए मूल (3) और \(-\frac{2}{5}\) हैं। परीक्षा में रैखिक गुणनखंडों के चिन्ह उलटकर मूल लिखें।
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Question 5/25
Medium Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
\(x^2-3x-10=0\) के लिए निम्न में से कौनसा कथन सही है?
Which statement is correct for \(x^2-3x-10=0\)?
#quadratic
#roots
#mixed-signs
A मूल (5) और (-2) हैं / The roots are (5) and (-2)
B मूल (-5) और (2) हैं / The roots are (-5) and (2)
C मूल (3) और (10) हैं / The roots are (3) and (10)
D मूल (-3) और (-10) हैं / The roots are (-3) and (-10)
Explanation opens after your attempt
Correct Answer
A. मूल (5) और (-2) हैं / The roots are (5) and (-2)
Step 1
Concept
(x-2 -3x-10=(x-5)(x+2)), so the roots are (5) and (-2). In exams, in mixed signs the larger value decides the middle-term sign.
Step 2
Why this answer is correct
The correct answer is A. मूल (5) और (-2) हैं / The roots are (5) and (-2). (x-2 -3x-10=(x-5)(x+2)), so the roots are (5) and (-2). In exams, in mixed signs the larger value decides the middle-term sign.
Step 3
Exam Tip
(x-2 -3x-10=(x-5)(x+2)), इसलिए मूल (5) और (-2) हैं। परीक्षा में मिश्रित चिन्हों में बड़ा मान मध्य पद का चिन्ह तय करता है।
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Question 6/25
Medium Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
\(x^2-2x-15=0\) को हल करने के लिए सही गुणनखंड रूप क्या है?
What is the correct factorised form to solve \(x^2-2x-15=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x-5)(x+3)=0)
B ((x+5)(x-3)=0)
C ((x-15)(x+1)=0)
D ((x+15)(x-1)=0)
Explanation opens after your attempt
Correct Answer
A. ((x-5)(x+3)=0)
Step 1
Concept
(-5+3=-2) and \(-5\times3=-15\), so ((x-5)(x+3)) is correct. In exams, match both product and sum.
Step 2
Why this answer is correct
The correct answer is A. ((x-5)(x+3)=0). (-5+3=-2) and \(-5\times3=-15\), so ((x-5)(x+3)) is correct. In exams, match both product and sum.
Step 3
Exam Tip
(-5+3=-2) और \(-5\times3=-15\), इसलिए ((x-5)(x+3)) सही है। परीक्षा में गुणनफल और योग दोनों मिलाएं।
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Question 7/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2+6x-55=0\) के मूल क्या हैं?
What are the roots of \(x^2+6x-55=0\)?
#quadratic
#roots
#mixed-signs
A (x=5,-11)
B (x=-5,11)
C (x=55,-1)
D (x=-55,1)
Explanation opens after your attempt
Correct Answer
A. (x=5,-11)
Step 1
Concept
((x+11)(x-5)=0), so (x=-11) and (x=5). In exams, the sign changes while finding roots from factors.
Step 2
Why this answer is correct
The correct answer is A. (x=5,-11). ((x+11)(x-5)=0), so (x=-11) and (x=5). In exams, the sign changes while finding roots from factors.
Step 3
Exam Tip
((x+11)(x-5)=0), इसलिए (x=-11) और (x=5) हैं। परीक्षा में गुणनखंड से मूल निकालते समय चिन्ह बदलता है।
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Question 8/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2+6x-55=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2+6x-55=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x+11)(x-5))
B ((x-11)(x+5))
C ((x+55)(x-1))
D ((x-55)(x+1))
Explanation opens after your attempt
Correct Answer
A. ((x+11)(x-5))
Step 1
Concept
(11+(-5)=6) and \(11\times(-5)=-55\), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.
Step 2
Why this answer is correct
The correct answer is A. ((x+11)(x-5)). (11+(-5)=6) and \(11\times(-5)=-55\), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.
Step 3
Exam Tip
(11+(-5)=6) और \(11\times(-5)=-55\), इसलिए ((x+11)(x-5)) सही है। परीक्षा में एक चिन्ह धनात्मक और एक ऋणात्मक रखें।
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Question 9/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 36
\(x^2-6x-27=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2-6x-27=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x-9)(x+3))
B ((x+9)(x-3))
C ((x-27)(x+1))
D ((x+9)(x+3))
Explanation opens after your attempt
Correct Answer
A. ((x-9)(x+3))
Step 1
Concept
Since (-9+3=-6) and \(-9\times3=-27\), ((x-9)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 2
Why this answer is correct
The correct answer is A. ((x-9)(x+3)). Since (-9+3=-6) and \(-9\times3=-27\), ((x-9)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 3
Exam Tip
(-9+3=-6) और \(-9\times3=-27\), इसलिए ((x-9)(x+3)) सही है। परीक्षा में मिश्रित चिन्ह वाले गुणनखंड ध्यान से चुनें।
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Question 10/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
\(x^2+5x-14=0\) के मूल क्या हैं?
What are the roots of \(x^2+5x-14=0\)?
#quadratic
#roots
#mixed-signs
A (x=2,-7)
B (x=-2,7)
C (x=14,-1)
D (x=-14,1)
Explanation opens after your attempt
Correct Answer
A. (x=2,-7)
Step 1
Concept
((x+7)(x-2)=0), so (x=-7) and (x=2). In exams, the sign changes while finding roots from factors.
Step 2
Why this answer is correct
The correct answer is A. (x=2,-7). ((x+7)(x-2)=0), so (x=-7) and (x=2). In exams, the sign changes while finding roots from factors.
Step 3
Exam Tip
((x+7)(x-2)=0), इसलिए (x=-7) और (x=2) हैं। परीक्षा में गुणनखंड से मूल निकालते समय चिन्ह बदलता है।
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Question 11/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
\(x^2+5x-14=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2+5x-14=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x+7)(x-2))
B ((x-7)(x+2))
C ((x+14)(x-1))
D ((x-14)(x+1))
Explanation opens after your attempt
Correct Answer
A. ((x+7)(x-2))
Step 1
Concept
(7+(-2)=5) and \(7\times(-2)=-14\), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.
Step 2
Why this answer is correct
The correct answer is A. ((x+7)(x-2)). (7+(-2)=5) and \(7\times(-2)=-14\), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.
Step 3
Exam Tip
(7+(-2)=5) और \(7\times(-2)=-14\), इसलिए ((x+7)(x-2)) सही है। परीक्षा में एक चिन्ह धनात्मक और एक ऋणात्मक रखें।
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Question 12/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 35
\(x^2-4x-21=0\) के सही गुणनखंड कौनसे हैं?
What are the correct factors of \(x^2-4x-21=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x-7)(x+3))
B ((x+7)(x-3))
C ((x-21)(x+1))
D ((x+7)(x+3))
Explanation opens after your attempt
Correct Answer
A. ((x-7)(x+3))
Step 1
Concept
Since (-7+3=-4) and \(-7\times3=-21\), ((x-7)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 2
Why this answer is correct
The correct answer is A. ((x-7)(x+3)). Since (-7+3=-4) and \(-7\times3=-21\), ((x-7)(x+3)) is correct. In exams, choose mixed signs carefully.
Step 3
Exam Tip
(-7+3=-4) और \(-7\times3=-21\), इसलिए ((x-7)(x+3)) सही है। परीक्षा में मिश्रित चिन्ह वाले गुणनखंड ध्यान से चुनें।
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Question 13/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
\(x^2+x-6=0\) के मूल क्या हैं?
What are the roots of \(x^2+x-6=0\)?
#quadratic
#roots
#mixed-signs
A (x=2,-3)
B (x=-2,3)
C (x=6,-1)
D (x=-6,1)
Explanation opens after your attempt
Correct Answer
A. (x=2,-3)
Step 1
Concept
((x+3)(x-2)=0), so (x=-3) and (x=2). In exams, the sign changes while finding roots from factors.
Step 2
Why this answer is correct
The correct answer is A. (x=2,-3). ((x+3)(x-2)=0), so (x=-3) and (x=2). In exams, the sign changes while finding roots from factors.
Step 3
Exam Tip
((x+3)(x-2)=0), इसलिए (x=-3) और (x=2) हैं। परीक्षा में गुणनखंड से मूल निकालते समय संकेत बदलता है।
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Question 14/25
Easy Mathematics
Quadratic Equations Methods of Solving Quadratic Equations Class 10 Level 34
\(x^2+x-6=0\) के लिए सही गुणनखंड कौनसे हैं?
What are the correct factors for \(x^2+x-6=0\)?
#quadratic
#factorisation
#mixed-signs
A ((x+3)(x-2))
B ((x-3)(x+2))
C ((x+6)(x-1))
D ((x-6)(x+1))
Explanation opens after your attempt
Correct Answer
A. ((x+3)(x-2))
Step 1
Concept
(3+(-2)=1) and \(3\times(-2)=-6\), so ((x+3)(x-2)) is correct. In exams, match signs with the product.
Step 2
Why this answer is correct
The correct answer is A. ((x+3)(x-2)). (3+(-2)=1) and \(3\times(-2)=-6\), so ((x+3)(x-2)) is correct. In exams, match signs with the product.
Step 3
Exam Tip
(3+(-2)=1) और \(3\times(-2)=-6\), इसलिए ((x+3)(x-2)) सही है। परीक्षा में चिन्हों को गुणनफल से मिलाएं।
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Question 15/25
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(x^2+2x-35=0\) के मूल कौन से हैं?
What are the roots of \(x^2+2x-35=0\)?
#roots
#factorisation
#mixed_signs
A (5) और (-7) / (5) and (-7)
B (7) और (-5) / (7) and (-5)
C (5) और (7) / (5) and (7)
D (-5) और (-7) / (-5) and (-7)
Explanation opens after your attempt
Correct Answer
A. (5) और (-7) / (5) and (-7)
Step 1
Concept
(x-2 +2x-35=(x+7)(x-5)). Therefore the roots are (-7) and (5).
Step 2
Why this answer is correct
The correct answer is A. (5) और (-7) / (5) and (-7). (x-2 +2x-35=(x+7)(x-5)). Therefore the roots are (-7) and (5).
Step 3
Exam Tip
(x-2 +2x-35=(x+7)(x-5)) है। इसलिए मूल (-7) और (5) हैं।
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Question 16/25
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 17/25
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2+x-20=0\) के मूल कौन से हैं?
What are the roots of \(x^2+x-20=0\)?
#roots
#factorisation
#mixed_signs
A (4) और (-5) / (4) and (-5)
B (5) और (-4) / (5) and (-4)
C (4) और (5) / (4) and (5)
D (-4) और (-5) / (-4) and (-5)
Explanation opens after your attempt
Correct Answer
A. (4) और (-5) / (4) and (-5)
Step 1
Concept
(x-2 +x-20=(x+5)(x-4)). Therefore the roots are (-5) and (4).
Step 2
Why this answer is correct
The correct answer is A. (4) और (-5) / (4) and (-5). (x-2 +x-20=(x+5)(x-4)). Therefore the roots are (-5) and (4).
Step 3
Exam Tip
(x-2 +x-20=(x+5)(x-4)) है। इसलिए मूल (-5) और (4) हैं।
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Question 18/25
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 19/25
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(x^2-2x-15=0\) के मूल कौन से हैं?
What are the roots of \(x^2-2x-15=0\)?
#roots
#factorisation
#mixed_signs
A (5) और (-3) / (5) and (-3)
B (3) और (-5) / (3) and (-5)
C (5) और (3) / (5) and (3)
D (-5) और (-3) / (-5) and (-3)
Explanation opens after your attempt
Correct Answer
A. (5) और (-3) / (5) and (-3)
Step 1
Concept
(x-2 -2x-15=(x-5)(x+3)). Therefore the roots are (5) and (-3).
Step 2
Why this answer is correct
The correct answer is A. (5) और (-3) / (5) and (-3). (x-2 -2x-15=(x-5)(x+3)). Therefore the roots are (5) and (-3).
Step 3
Exam Tip
(x-2 -2x-15=(x-5)(x+3)) है। इसलिए मूल (5) और (-3) हैं।
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Question 20/25
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(x^2+x-12=0\) के मूल कौन से हैं?
What are the roots of \(x^2+x-12=0\)?
#roots
#factorisation
#mixed_signs
A (3) और (-4) / (3) and (-4)
B (4) और (-3) / (4) and (-3)
C (-3) और (-4) / (-3) and (-4)
D (3) और (4) / (3) and (4)
Explanation opens after your attempt
Correct Answer
A. (3) और (-4) / (3) and (-4)
Step 1
Concept
(x-2 +x-12=(x+4)(x-3)). Therefore the roots are (-4) and (3).
Step 2
Why this answer is correct
The correct answer is A. (3) और (-4) / (3) and (-4). (x-2 +x-12=(x+4)(x-3)). Therefore the roots are (-4) and (3).
Step 3
Exam Tip
(x-2 +x-12=(x+4)(x-3)) है। इसलिए मूल (-4) और (3) हैं।
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Question 21/25
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 22/25
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2-3x-10=0\) के मूल कौन से हैं?
What are the roots of \(x^2-3x-10=0\)?
#roots
#factorisation
#mixed_signs
A (5) और (-2) / (5) and (-2)
B (2) और (-5) / (2) and (-5)
C (-5) और (-2) / (-5) and (-2)
D (5) और (2) / (5) and (2)
Explanation opens after your attempt
Correct Answer
A. (5) और (-2) / (5) and (-2)
Step 1
Concept
(x-2 -3x-10=(x-5)(x+2)). Therefore the roots are (5) and (-2).
Step 2
Why this answer is correct
The correct answer is A. (5) और (-2) / (5) and (-2). (x-2 -3x-10=(x-5)(x+2)). Therefore the roots are (5) and (-2).
Step 3
Exam Tip
(x-2 -3x-10=(x-5)(x+2)) है। इसलिए मूल (5) और (-2) हैं।
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Question 23/25
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 24/25
Easy Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(x^2+2x-8=0\) के मूल कौन से हैं?
What are the roots of \(x^2+2x-8=0\)?
#roots
#factorisation
#mixed_signs
A (4) और (-2) / (4) and (-2)
B (2) और (-4) / (2) and (-4)
C (-2) और (-4) / (-2) and (-4)
D (2) और (4) / (2) and (4)
Explanation opens after your attempt
Correct Answer
B. (2) और (-4) / (2) and (-4)
Step 1
Concept
(x-2 +2x-8=(x+4)(x-2)) so the roots are (-4) and (2). Roots have opposite signs to the factor constants.
Step 2
Why this answer is correct
The correct answer is B. (2) और (-4) / (2) and (-4). (x-2 +2x-8=(x+4)(x-2)) so the roots are (-4) and (2). Roots have opposite signs to the factor constants.
Step 3
Exam Tip
(x-2 +2x-8=(x+4)(x-2)) इसलिए मूल (-4) और (2) हैं। गुणनखंड के चिन्ह उलटकर मूल मिलते हैं।
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Question 25/25
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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