किसी द्विघात समीकरण का मूल किसे कहते हैं?
What is called a root of a quadratic equation?
#roots
#definition
#concept
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A वह मान जिससे समीकरण सत्य हो / The value that satisfies the equation
B केवल सबसे बड़ा गुणांक / Only the greatest coefficient
C केवल अचर पद / Only the constant term
D समीकरण की घात / The degree of the equation
Explanation opens after your attempt
Correct Answer
A. वह मान जिससे समीकरण सत्य हो / The value that satisfies the equation
Step 1
Concept
A root is a value that makes the equation equal to (0). In exams first substitute the value and check.
Step 2
Why this answer is correct
The correct answer is A. वह मान जिससे समीकरण सत्य हो / The value that satisfies the equation. A root is a value that makes the equation equal to (0). In exams first substitute the value and check.
Step 3
Exam Tip
मूल वह मान है जिसे रखने पर समीकरण का मान (0) हो जाता है। परीक्षा में पहले मान रखकर जांचें।
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यदि (p(2)=0) है तो (2) के बारे में सही कथन कौन सा है?
If (p(2)=0) then which statement about (2) is correct?
#roots
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#concept
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A (2) (p(x)=0) का मूल है / (2) is a root of (p(x)=0)
B (2) गुणांक है / (2) is a coefficient
C (2) अचर पद है / (2) is a constant term
D (2) घात है / (2) is the degree
Explanation opens after your attempt
Correct Answer
A. (2) (p(x)=0) का मूल है / (2) is a root of (p(x)=0)
Step 1
Concept
(p(2)=0) means the equation is satisfied when (x=2). Substitution is the key method in such questions.
Step 2
Why this answer is correct
The correct answer is A. (2) (p(x)=0) का मूल है / (2) is a root of (p(x)=0). (p(2)=0) means the equation is satisfied when (x=2). Substitution is the key method in such questions.
Step 3
Exam Tip
(p(2)=0) का अर्थ है कि (x=2) रखने पर समीकरण संतुष्ट होता है। ऐसे प्रश्न में प्रतिस्थापन मुख्य तरीका है।
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क्या (x=2) समीकरण \(x^2-5x+6=0\) का मूल है?
Is (x=2) a root of \(x^2-5x+6=0\)?
#roots
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#numerical
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A हाँ / Yes
B नहीं / No
C केवल (x=0) पर / Only at (x=0)
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=2) gives (4-10+6=0) so it is a root. In exams always check the final sum after substitution.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=2) gives (4-10+6=0) so it is a root. In exams always check the final sum after substitution.
Step 3
Exam Tip
(x=2) रखने पर (4-10+6=0) मिलता है इसलिए यह मूल है। परीक्षा में मान रखने के बाद अंतिम योग जरूर देखें।
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समीकरण \(x^2-9=0\) के मूल कौन से हैं?
What are the roots of \(x^2-9=0\)?
#roots
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#difference_of_squares
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A (3) और (-3) / (3) and (-3)
B (9) और (-9) / (9) and (-9)
C (0) और (9) / (0) and (9)
D केवल (3) / Only (3)
Explanation opens after your attempt
Correct Answer
A. (3) और (-3) / (3) and (-3)
Step 1
Concept
(x-2 -9=(x-3)(x+3)) so the roots are (3) and (-3). In exams quickly identify the difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (3) और (-3) / (3) and (-3). (x-2 -9=(x-3)(x+3)) so the roots are (3) and (-3). In exams quickly identify the difference of squares.
Step 3
Exam Tip
(x-2 -9=(x-3)(x+3)) इसलिए मूल (3) और (-3) हैं। परीक्षा में वर्गों के अंतर को तुरंत पहचानें।
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समीकरण \(x^2-4x+3=0\) के मूल कौन से हैं?
What are the roots of \(x^2-4x+3=0\)?
#roots
#factorisation
#quadratic
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A (1) और (3) / (1) and (3)
B (-1) और (-3) / (-1) and (-3)
C (2) और (2) / (2) and (2)
D (0) और (3) / (0) and (3)
Explanation opens after your attempt
Correct Answer
A. (1) और (3) / (1) and (3)
Step 1
Concept
(x-2 -4x+3=(x-1)(x-3)) so the roots are (1) and (3). Factorisation is the fastest method for easy questions.
Step 2
Why this answer is correct
The correct answer is A. (1) और (3) / (1) and (3). (x-2 -4x+3=(x-1)(x-3)) so the roots are (1) and (3). Factorisation is the fastest method for easy questions.
Step 3
Exam Tip
(x-2 -4x+3=(x-1)(x-3)) इसलिए मूल (1) और (3) हैं। आसान प्रश्न में गुणनखंड बनाना सबसे तेज तरीका है।
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समीकरण \(x^2+5x+6=0\) के मूल कौन से हैं?
What are the roots of \(x^2+5x+6=0\)?
#roots
#factorisation
#signs
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A (-2) और (-3) / (-2) and (-3)
B (2) और (3) / (2) and (3)
C (-1) और (-6) / (-1) and (-6)
D (0) और (-5) / (0) and (-5)
Explanation opens after your attempt
Correct Answer
A. (-2) और (-3) / (-2) and (-3)
Step 1
Concept
(x-2 +5x+6=(x+2)(x+3)) so the roots are (-2) and (-3). Pay special attention to signs.
Step 2
Why this answer is correct
The correct answer is A. (-2) और (-3) / (-2) and (-3). (x-2 +5x+6=(x+2)(x+3)) so the roots are (-2) and (-3). Pay special attention to signs.
Step 3
Exam Tip
(x-2 +5x+6=(x+2)(x+3)) इसलिए मूल (-2) और (-3) हैं। चिन्हों पर विशेष ध्यान दें।
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किस समीकरण के मूल (0) और (5) हैं?
Which equation has roots (0) and (5)?
#roots
#equation_from_roots
#zero_root
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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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यदि (1) समीकरण \(x^2-kx+2=0\) का मूल है तो (k) का मान क्या है?
If (1) is a root of \(x^2-kx+2=0\) then what is the value of (k)?
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#parameter
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A (3)
B (2)
C (1)
D (-3)
Explanation opens after your attempt
Step 1
Concept
Putting (x=1) gives (1-k+2=0) so (k=3). Substitute the given root directly in the equation.
Step 2
Why this answer is correct
The correct answer is A. (3). Putting (x=1) gives (1-k+2=0) so (k=3). Substitute the given root directly in the equation.
Step 3
Exam Tip
(x=1) रखने पर (1-k+2=0) इसलिए (k=3) है। दिए गए मूल को सीधे समीकरण में रखें।
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समीकरण \(ax^2+bx+c=0\) के मूल \(\alpha\) और \(\beta\) हों तो \(\alpha+\beta\) किसके बराबर होता है?
If \(\alpha\) and \(\beta\) are roots of \(ax^2+bx+c=0\) then what is \(\alpha+\beta\)?
#roots
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#formula
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A \(-\frac{b}{a}\)
B \(\frac{c}{a}\)
C \(\frac{b}{a}\)
D \(-\frac{c}{a}\)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{b}{a}\)
Step 1
Concept
For a quadratic equation the sum of roots is \(-\frac{b}{a}\). Keep the sign of (b) correct while using the formula.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{b}{a}\). For a quadratic equation the sum of roots is \(-\frac{b}{a}\). Keep the sign of (b) correct while using the formula.
Step 3
Exam Tip
द्विघात समीकरण में मूलों का योग \(-\frac{b}{a}\) होता है। सूत्र लगाते समय (b) का चिन्ह सही रखें।
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समीकरण \(ax^2+bx+c=0\) के मूल \(\alpha\) और \(\beta\) हों तो \(\alpha\beta\) किसके बराबर होता है?
If \(\alpha\) and \(\beta\) are roots of \(ax^2+bx+c=0\) then what is \(\alpha\beta\)?
#roots
#product_of_roots
#formula
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A \(\frac{c}{a}\)
B \(-\frac{b}{a}\)
C \(\frac{b}{c}\)
D \(-\frac{a}{c}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{c}{a}\)
Step 1
Concept
For a quadratic equation the product of roots is \(\frac{c}{a}\). Do not ignore (a).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{c}{a}\). For a quadratic equation the product of roots is \(\frac{c}{a}\). Do not ignore (a).
Step 3
Exam Tip
द्विघात समीकरण में मूलों का गुणनफल \(\frac{c}{a}\) होता है। (a) को नजरअंदाज न करें।
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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
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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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समीकरण \(x^2-4x+4=0\) के डिस्क्रिमिनेंट (D) का मान क्या है?
What is the value of discriminant (D) for \(x^2-4x+4=0\)?
#roots
#discriminant
#calculation
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A (4)
B (0)
C (-4)
D (8)
Explanation opens after your attempt
Step 1
Concept
Here \(D=b^2-4ac=16-16=0\). Therefore its roots will be equal.
Step 2
Why this answer is correct
The correct answer is B. (0). Here \(D=b^2-4ac=16-16=0\). Therefore its roots will be equal.
Step 3
Exam Tip
यहां \(D=b^2-4ac=16-16=0\) है। इसलिए इसके मूल बराबर होंगे।
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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
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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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समीकरण \(x^2+1=0\) के वास्तविक मूलों के बारे में सही कथन कौन सा है?
Which statement is correct about the real roots of \(x^2+1=0\)?
#roots
#no_real_roots
#concept
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A दो वास्तविक मूल हैं / It has two real roots
B एक वास्तविक मूल है / It has one real root
C कोई वास्तविक मूल नहीं है / It has no real roots
D मूल (1) और (-1) हैं / The roots are (1) and (-1)
Explanation opens after your attempt
Correct Answer
C. कोई वास्तविक मूल नहीं है / It has no real roots
Step 1
Concept
\(x^2\) is never negative so \(x^2+1=0\) is not satisfied by any real number. Sign checking helps solve such questions quickly.
Step 2
Why this answer is correct
The correct answer is C. कोई वास्तविक मूल नहीं है / It has no real roots. \(x^2\) is never negative so \(x^2+1=0\) is not satisfied by any real number. Sign checking helps solve such questions quickly.
Step 3
Exam Tip
\(x^2\) कभी ऋणात्मक नहीं होता इसलिए \(x^2+1=0\) वास्तविक संख्या से संतुष्ट नहीं होता। संकेत से ऐसे प्रश्न जल्दी हल होते हैं।
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क्या (x=0) समीकरण \(3x^2+2x=0\) का मूल है?
Is (x=0) a root of \(3x^2+2x=0\)?
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#checking
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A हाँ / Yes
B नहीं / No
C केवल जब (x=2) हो / Only when (x=2)
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. हाँ / Yes
Step 1
Concept
Putting (x=0) gives (3(0)2 +2(0)=0). Therefore (0) is a root.
Step 2
Why this answer is correct
The correct answer is A. हाँ / Yes. Putting (x=0) gives (3(0)2 +2(0)=0). Therefore (0) is a root.
Step 3
Exam Tip
(x=0) रखने पर (3(0)2 +2(0)=0) मिलता है। इसलिए (0) मूल है।
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समीकरण \(x^2=x\) के मूल कौन से हैं?
What are the roots of \(x^2=x\)?
#roots
#factorisation
#simple_equation
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A (0) और (1) / (0) and (1)
B (1) और (2) / (1) and (2)
C (-1) और (1) / (-1) and (1)
D (0) और (-1) / (0) and (-1)
Explanation opens after your attempt
Correct Answer
A. (0) और (1) / (0) and (1)
Step 1
Concept
Writing \(x^2=x\) as \(x^2-x=0\) gives (x(x-1)=0). So the roots are (0) and (1).
Step 2
Why this answer is correct
The correct answer is A. (0) और (1) / (0) and (1). Writing \(x^2=x\) as \(x^2-x=0\) gives (x(x-1)=0). So the roots are (0) and (1).
Step 3
Exam Tip
\(x^2=x\) को \(x^2-x=0\) लिखने पर (x(x-1)=0) मिलता है। इसलिए मूल (0) और (1) हैं।
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समीकरण \(2x^2-8=0\) के मूल कौन से हैं?
What are the roots of \(2x^2-8=0\)?
#roots
#square_root
#plus_minus
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A (2) और (-2) / (2) and (-2)
B (4) और (-4) / (4) and (-4)
C (0) और (2) / (0) and (2)
D केवल (2) / Only (2)
Explanation opens after your attempt
Correct Answer
A. (2) और (-2) / (2) and (-2)
Step 1
Concept
From \(2x^2-8=0\) we get \(x^2=4\) so \(x=\pm 2\). Take both signs while finding square roots.
Step 2
Why this answer is correct
The correct answer is A. (2) और (-2) / (2) and (-2). From \(2x^2-8=0\) we get \(x^2=4\) so \(x=\pm 2\). Take both signs while finding square roots.
Step 3
Exam Tip
\(2x^2-8=0\) से \(x^2=4\) मिलता है इसलिए \(x=\pm 2\)। वर्गमूल लेते समय दोनों चिन्ह लें।
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समीकरण \(x^2+2x+1=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+2x+1=0\)?
#roots
#repeated_root
#perfect_square
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A (1)
B (-1)
C (2)
D (-2)
Explanation opens after your attempt
Step 1
Concept
(x-2 +2x+1=(x+1)2 ) so the repeated root is (-1). Recognising a perfect square is useful.
Step 2
Why this answer is correct
The correct answer is B. (-1). (x-2 +2x+1=(x+1)2 ) so the repeated root is (-1). Recognising a perfect square is useful.
Step 3
Exam Tip
(x-2 +2x+1=(x+1)2 ) इसलिए दोहराया मूल (-1) है। पूर्ण वर्ग को पहचानना उपयोगी 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
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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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समीकरण \(x^2-6x+8=0\) का एक मूल (4) है तो दूसरा मूल क्या है?
If one root of \(x^2-6x+8=0\) is (4) then what is the other root?
#roots
#other_root
#factorisation
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A (1)
B (2)
C (-2)
D (8)
Explanation opens after your attempt
Step 1
Concept
(x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 2
Why this answer is correct
The correct answer is B. (2). (x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 3
Exam Tip
(x-2 -6x+8=(x-4)(x-2)) इसलिए दूसरा मूल (2) है। दिए गए एक मूल से दूसरा गुणनखंड खोजें।
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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
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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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यदि (x-a) किसी द्विघात बहुपद का गुणनखंड है तो (a) क्या होगा?
If (x-a) is a factor of a quadratic polynomial then what is (a)?
#roots
#factor_theorem
#concept
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A गुणांक / Coefficient
B मूल / Root
C घात / Degree
D अचर पद / Constant term
Explanation opens after your attempt
Correct Answer
B. मूल / Root
Step 1
Concept
If (x-a) is a factor then substituting (x=a) makes the value (0). So (a) is a root.
Step 2
Why this answer is correct
The correct answer is B. मूल / Root. If (x-a) is a factor then substituting (x=a) makes the value (0). So (a) is a root.
Step 3
Exam Tip
गुणनखंड (x-a) होने पर (x=a) रखने से मान (0) होता है। इसलिए (a) मूल है।
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यदि (3) समीकरण \(x^2+px-6=0\) का मूल है तो (p) का मान क्या है?
If (3) is a root of \(x^2+px-6=0\) then what is the value of (p)?
#roots
#parameter
#checking
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A (1)
B (-1)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) gives (9+3p-6=0) so (p=-1). For a parameter question substitute the root directly.
Step 2
Why this answer is correct
The correct answer is B. (-1). Putting (x=3) gives (9+3p-6=0) so (p=-1). For a parameter question substitute the root directly.
Step 3
Exam Tip
(x=3) रखने पर (9+3p-6=0) इसलिए (p=-1)। पैरामीटर के लिए मूल को सीधे रखें।
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समीकरण \(5x^2=0\) के मूल के बारे में सही कथन कौन सा है?
Which statement is correct about the root of \(5x^2=0\)?
#roots
#repeated_zero
#concept
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A (0) दोहराया हुआ मूल है / (0) is a repeated root
B (5) दोहराया हुआ मूल है / (5) is a repeated root
C (1) और (5) मूल हैं / (1) and (5) are roots
D कोई वास्तविक मूल नहीं है / It has no real root
Explanation opens after your attempt
Correct Answer
A. (0) दोहराया हुआ मूल है / (0) is a repeated root
Step 1
Concept
From \(5x^2=0\) we get \(x^2=0\) so the root (0) occurs twice. For a zero root also check the factor.
Step 2
Why this answer is correct
The correct answer is A. (0) दोहराया हुआ मूल है / (0) is a repeated root. From \(5x^2=0\) we get \(x^2=0\) so the root (0) occurs twice. For a zero root also check the factor.
Step 3
Exam Tip
\(5x^2=0\) से \(x^2=0\) मिलता है इसलिए मूल (0) दो बार आता है। शून्य मूल में भी गुणनखंड देखें।
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निम्न में से कौन सा मान \(x^2-2x=0\) का मूल है?
Which of the following values is a root of \(x^2-2x=0\)?
#roots
#select_root
#factorisation
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A (3)
B (-2)
C (2)
D (4)
Explanation opens after your attempt
Step 1
Concept
(x-2 -2x=x(x-2)) so the roots are (0) and (2). Choose the root that appears in the options.
Step 2
Why this answer is correct
The correct answer is C. (2). (x-2 -2x=x(x-2)) so the roots are (0) and (2). Choose the root that appears in the options.
Step 3
Exam Tip
(x-2 -2x=x(x-2)) इसलिए मूल (0) और (2) हैं। विकल्पों में मौजूद मूल को चुनें।
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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
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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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समीकरण \(x^2+7x+10=0\) के मूलों का योग क्या है?
What is the sum of roots of \(x^2+7x+10=0\)?
#roots
#sum_of_roots
#numerical
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A (7)
B (10)
C (-7)
D (-10)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is \(-\frac{b}{a}\) so here it is (-7). This can be found quickly without factorising.
Step 2
Why this answer is correct
The correct answer is C. (-7). The sum of roots is \(-\frac{b}{a}\) so here it is (-7). This can be found quickly without factorising.
Step 3
Exam Tip
मूलों का योग \(-\frac{b}{a}\) है इसलिए यहां (-7) होगा। गुणनखंड न बनाकर भी यह जल्दी निकलता है।
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समीकरण \(2x^2-3x-2=0\) के मूलों का गुणनफल क्या है?
What is the product of roots of \(2x^2-3x-2=0\)?
#roots
#product_of_roots
#numerical
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A (-1)
B (1)
C (-2)
D (2)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{c}{a}\) so \(\frac{-2}{2}=-1\). Identify (a) and (c) first.
Step 2
Why this answer is correct
The correct answer is A. (-1). The product of roots is \(\frac{c}{a}\) so \(\frac{-2}{2}=-1\). Identify (a) and (c) first.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}\) है इसलिए \(\frac{-2}{2}=-1\)। पहले (a) और (c) पहचानें।
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किस शर्त पर (0) समीकरण \(ax^2+bx+c=0\) का मूल होगा?
Under which condition will (0) be a root of \(ax^2+bx+c=0\)?
#roots
#zero_root
#condition
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A (c=0)
B (a=0)
C (b=0)
D (a=b)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) makes the equation (c=0). So for a zero root the constant term must be zero.
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) makes the equation (c=0). So for a zero root the constant term must be zero.
Step 3
Exam Tip
(x=0) रखने पर समीकरण (c=0) बनता है। इसलिए शून्य मूल के लिए अचर पद शून्य होना चाहिए।
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यदि (-2) समीकरण \(x^2+kx+4=0\) का मूल है तो (k) का मान क्या है?
If (-2) is a root of \(x^2+kx+4=0\) then what is the value of (k)?
#roots
#parameter
#negative_root
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A (2)
B (4)
C (-4)
D (-2)
Explanation opens after your attempt
Step 1
Concept
Putting (x=-2) gives (4-2k+4=0) so (k=4). Be careful with signs while substituting a negative root.
Step 2
Why this answer is correct
The correct answer is B. (4). Putting (x=-2) gives (4-2k+4=0) so (k=4). Be careful with signs while substituting a negative root.
Step 3
Exam Tip
(x=-2) रखने पर (4-2k+4=0) इसलिए (k=4)। ऋणात्मक मूल रखते समय चिन्ह सावधानी से लगाएं।
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समीकरण \(x^2+6x+9=0\) के मूल कौन से हैं?
What are the roots of \(x^2+6x+9=0\)?
#roots
#repeated_root
#perfect_square
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A (3) और (3) / (3) and (3)
B (-3) और (-3) / (-3) and (-3)
C (0) और (9) / (0) and (9)
D (-6) और (-9) / (-6) and (-9)
Explanation opens after your attempt
Correct Answer
B. (-3) और (-3) / (-3) and (-3)
Step 1
Concept
(x-2 +6x+9=(x+3)2 ) so both roots are (-3). In a perfect square the root sign is opposite.
Step 2
Why this answer is correct
The correct answer is B. (-3) और (-3) / (-3) and (-3). (x-2 +6x+9=(x+3)2 ) so both roots are (-3). In a perfect square the root sign is opposite.
Step 3
Exam Tip
(x-2 +6x+9=(x+3)2 ) इसलिए दोनों मूल (-3) हैं। पूर्ण वर्ग में मूल का चिन्ह उल्टा होता है।
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समीकरण \(3x^2-12x=0\) के मूल कौन से हैं?
What are the roots of \(3x^2-12x=0\)?
#roots
#common_factor
#factorisation
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A (0) और (4) / (0) and (4)
B (3) और (4) / (3) and (4)
C (-4) और (0) / (-4) and (0)
D (0) और (12) / (0) and (12)
Explanation opens after your attempt
Correct Answer
A. (0) और (4) / (0) and (4)
Step 1
Concept
(3x-2 -12x=3x(x-4)) so the roots are (0) and (4). Take out the common factor first.
Step 2
Why this answer is correct
The correct answer is A. (0) और (4) / (0) and (4). (3x-2 -12x=3x(x-4)) so the roots are (0) and (4). Take out the common factor first.
Step 3
Exam Tip
(3x-2 -12x=3x(x-4)) इसलिए मूल (0) और (4) हैं। सामान्य गुणनखंड पहले बाहर निकालें।
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समीकरण \(x^2-10x+25=0\) का मूल कौन सा है?
Which is the root of \(x^2-10x+25=0\)?
#roots
#repeated_root
#perfect_square
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A (-5) दो बार / (-5) twice
B (5) दो बार / (5) twice
C (10) दो बार / (10) twice
D (0) दो बार / (0) twice
Explanation opens after your attempt
Correct Answer
B. (5) दो बार / (5) twice
Step 1
Concept
(x-2 -10x+25=(x-5)2 ) so the repeated root is (5). Recognising a perfect square saves time.
Step 2
Why this answer is correct
The correct answer is B. (5) दो बार / (5) twice. (x-2 -10x+25=(x-5)2 ) so the repeated root is (5). Recognising a perfect square saves time.
Step 3
Exam Tip
(x-2 -10x+25=(x-5)2 ) इसलिए दोहराया मूल (5) है। पूर्ण वर्ग पहचानना समय बचाता है।
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निम्न में से कौन सा मान \(x^2-16=0\) का मूल नहीं है?
Which of the following values is not a root of \(x^2-16=0\)?
#roots
#not_a_root
#checking
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A (4)
B (-4)
C (0)
D \(\sqrt{16}\)
Explanation opens after your attempt
Step 1
Concept
The roots of \(x^2-16=0\) are (4) and (-4). Substituting (0) does not satisfy the equation.
Step 2
Why this answer is correct
The correct answer is C. (0). The roots of \(x^2-16=0\) are (4) and (-4). Substituting (0) does not satisfy the equation.
Step 3
Exam Tip
\(x^2-16=0\) के मूल (4) और (-4) हैं। (0) रखने पर समीकरण संतुष्ट नहीं होता।
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यदि ((x-2)(x+7)=0) है तो मूल कौन से होंगे?
If ((x-2)(x+7)=0) then what are the roots?
#roots
#zero_product_rule
#factor_form
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A (2) और (7) / (2) and (7)
B (-2) और (-7) / (-2) and (-7)
C (2) और (-7) / (2) and (-7)
D (-2) और (7) / (-2) and (7)
Explanation opens after your attempt
Correct Answer
C. (2) और (-7) / (2) and (-7)
Step 1
Concept
By the zero product rule (x-2=0) or (x+7=0). Hence the roots are (2) and (-7).
Step 2
Why this answer is correct
The correct answer is C. (2) और (-7) / (2) and (-7). By the zero product rule (x-2=0) or (x+7=0). Hence the roots are (2) and (-7).
Step 3
Exam Tip
शून्य गुणनफल नियम से (x-2=0) या (x+7=0) होगा। इसलिए मूल (2) और (-7) हैं।
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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
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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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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
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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 (1) समीकरण \(2x^2-3x+m=0\) का मूल है तो (m) का मान क्या है?
If (1) is a root of \(2x^2-3x+m=0\) then what is the value of (m)?
#roots
#parameter
#substitution
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A (-1)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
Putting (x=1) gives (2-3+m=0) so (m=1). When a root is given substitute directly in the equation.
Step 2
Why this answer is correct
The correct answer is B. (1). Putting (x=1) gives (2-3+m=0) so (m=1). When a root is given substitute directly in the equation.
Step 3
Exam Tip
(x=1) रखने पर (2-3+m=0) इसलिए (m=1)। मूल दिया हो तो समीकरण में सीधा प्रतिस्थापन करें।
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सूत्र या गुणनखंडन से \(x^2-3x+2=0\) के मूल कौन से हैं?
Using formula or factorisation what are the roots of \(x^2-3x+2=0\)?
#roots
#factorisation
#formula_check
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A (1) और (2) / (1) and (2)
B (-1) और (-2) / (-1) and (-2)
C (0) और (3) / (0) and (3)
D (2) और (3) / (2) and (3)
Explanation opens after your attempt
Correct Answer
A. (1) और (2) / (1) and (2)
Step 1
Concept
(x-2 -3x+2=(x-1)(x-2)) so the roots are (1) and (2). For small numbers factorisation is faster.
Step 2
Why this answer is correct
The correct answer is A. (1) और (2) / (1) and (2). (x-2 -3x+2=(x-1)(x-2)) so the roots are (1) and (2). For small numbers factorisation is faster.
Step 3
Exam Tip
(x-2 -3x+2=(x-1)(x-2)) इसलिए मूल (1) और (2) हैं। छोटे अंकों में गुणनखंडन तेज रहता है।
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यदि (2) किसी द्विघात बहुपद का मूल है तो कौन सा गुणनखंड निश्चित होगा?
If (2) is a root of a quadratic polynomial then which factor is certain?
#roots
#factor_from_root
#concept
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A (x+2)
B (x-2)
C (2x+1)
D \(x^2+2\)
Explanation opens after your attempt
Step 1
Concept
If the root is (2) the corresponding factor is (x-2). For root (r) remember the factor (x-r).
Step 2
Why this answer is correct
The correct answer is B. (x-2). If the root is (2) the corresponding factor is (x-2). For root (r) remember the factor (x-r).
Step 3
Exam Tip
मूल (2) होने पर संबंधित गुणनखंड (x-2) होता है। मूल (r) के लिए गुणनखंड (x-r) याद रखें।
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समीकरण \(x^2+4=0\) के वास्तविक मूलों की संख्या कितनी है?
How many real roots does \(x^2+4=0\) have?
#roots
#no_real_roots
#number_of_roots
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A (2)
B (1)
C (0)
D (4)
Explanation opens after your attempt
Step 1
Concept
Because \(x^2\ge 0\) so \(x^2+4\) never becomes (0). Therefore it has no real root.
Step 2
Why this answer is correct
The correct answer is C. (0). Because \(x^2\ge 0\) so \(x^2+4\) never becomes (0). Therefore it has no real root.
Step 3
Exam Tip
क्योंकि \(x^2\ge 0\) होता है इसलिए \(x^2+4\) कभी (0) नहीं बनता। इसलिए कोई वास्तविक मूल नहीं है।
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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
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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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समीकरण \(x^2=16\) के मूल कौन से हैं?
What are the roots of \(x^2=16\)?
#roots
#square_root
#plus_minus
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A (4) और (-4) / (4) and (-4)
B केवल (4) / Only (4)
C (8) और (-8) / (8) and (-8)
D (0) और (16) / (0) and (16)
Explanation opens after your attempt
Correct Answer
A. (4) और (-4) / (4) and (-4)
Step 1
Concept
From \(x^2=16\) we get \(x=\pm4\). In a square equation check both positive and negative roots.
Step 2
Why this answer is correct
The correct answer is A. (4) और (-4) / (4) and (-4). From \(x^2=16\) we get \(x=\pm4\). In a square equation check both positive and negative roots.
Step 3
Exam Tip
\(x^2=16\) से \(x=\pm4\) मिलता है। वर्ग समीकरण में धनात्मक और ऋणात्मक दोनों मूल देखें।
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यदि एक मूल (3) है और मूलों का गुणनफल (12) है तो दूसरा मूल क्या होगा?
If one root is (3) and the product of roots is (12) then what is the other root?
#roots
#other_root
#product
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A (3)
B (4)
C (9)
D (12)
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Step 1
Concept
The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 2
Why this answer is correct
The correct answer is B. (4). The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{12}{3}=4\) होगा। गुणनफल वाले प्रश्न में दिए मूल से भाग दें।
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यदि मूलों का योग (5) है और एक मूल (2) है तो दूसरा मूल क्या है?
If the sum of roots is (5) and one root is (2) then what is the other root?
#roots
#other_root
#sum
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A (2)
B (3)
C (5)
D (7)
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Step 1
Concept
The other root is (5-2=3). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (3). The other root is (5-2=3). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (5-2=3) है। योग में से दिए हुए मूल को घटाएं।
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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
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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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समीकरण \(2x^2-5x+3=0\) के मूल कौन से हैं?
What are the roots of \(2x^2-5x+3=0\)?
#roots
#factorisation
#coefficient
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A (1) और \(\frac{3}{2}\) / (1) and \(\frac{3}{2}\)
B (-1) और \(\frac{3}{2}\) / (-1) and \(\frac{3}{2}\)
C (2) और (3) / (2) and (3)
D \(\frac{1}{2}\) और (3) / \(\frac{1}{2}\) and (3)
Explanation opens after your attempt
Correct Answer
A. (1) और \(\frac{3}{2}\) / (1) and \(\frac{3}{2}\)
Step 1
Concept
(2x-2 -5x+3=(2x-3)(x-1)) so the roots are \(\frac{3}{2}\) and (1). Be careful with factors that contain coefficients.
Step 2
Why this answer is correct
The correct answer is A. (1) और \(\frac{3}{2}\) / (1) and \(\frac{3}{2}\). (2x-2 -5x+3=(2x-3)(x-1)) so the roots are \(\frac{3}{2}\) and (1). Be careful with factors that contain coefficients.
Step 3
Exam Tip
(2x-2 -5x+3=(2x-3)(x-1)) इसलिए मूल \(\frac{3}{2}\) और (1) हैं। गुणांक वाले गुणनखंडों में सावधानी रखें।
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निम्न में से किस समीकरण का एक मूल (0) है?
Which of the following equations has one root (0)?
#roots
#zero_root
#identification
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A \(x^2+5x=0\)
B \(x^2+5=0\)
C \(x^2-5x+6=0\)
D \(x^2+2x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+5x=0\)
Step 1
Concept
(x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+5x=0\). (x-2 +5x=x(x+5)) so (x=0) is one root. If the constant term is (0) check for a zero root.
Step 3
Exam Tip
(x-2 +5x=x(x+5)) इसलिए (x=0) एक मूल है। अचर पद (0) हो तो शून्य मूल की संभावना देखें।
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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
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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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यदि (x=3) समीकरण \(2x^2-7x+k=0\) का मूल है तो (k) का मान क्या होगा?
If (x=3) is a root of \(2x^2-7x+k=0\), what will be the value of (k)?
#roots
#parameter
#substitution
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A (3)
B (6)
C (-3)
D (-6)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) gives (18-21+k=0), so (k=3). In such questions substitute the given root directly into the equation.
Step 2
Why this answer is correct
The correct answer is A. (3). Putting (x=3) gives (18-21+k=0), so (k=3). In such questions substitute the given root directly into the equation.
Step 3
Exam Tip
(x=3) रखने पर (18-21+k=0) इसलिए (k=3) मिलता है। ऐसे प्रश्नों में दिए हुए मूल को सीधे समीकरण में रखें।
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