यदि \(\alpha,\beta\) समीकरण \(8x^2-31x+15=0\) के मूल हैं, तो \(\alpha\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(8x^2-31x+15=0\), what is \(\alpha\beta\)?
#quadratic
#product-of-roots
#formula
A \( \frac{15}{8}\)
B \( \frac{31}{8}\)
C (15)
D \(-\frac{15}{8}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{15}{8}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{15}{8}\). In exams, use \(\frac{c}{a}\) for the product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{15}{8}\). The product of roots is \(\frac{c}{a}=\frac{15}{8}\). In exams, use \(\frac{c}{a}\) for the product.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{15}{8}\) है। परीक्षा में गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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यदि \(10x^2-23x+p=0\) के मूलों का गुणनफल \(\frac{2}{5}\) है, तो (p) क्या होगा?
If the product of roots of \(10x^2-23x+p=0\) is \(\frac{2}{5}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (4)
B (10)
C \( \frac{1}{25}\)
D \( \frac{23}{5}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{10}\), so \(\frac{p}{10}=\frac{2}{5}\) gives (p=4). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (4). The product of roots is \(\frac{p}{10}\), so \(\frac{p}{10}=\frac{2}{5}\) gives (p=4). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{10}\) है, इसलिए \(\frac{p}{10}=\frac{2}{5}\) से (p=4) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो (q) का मान क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (150)
B (250)
C (15)
D (25)
Explanation opens after your attempt
Step 1
Concept
The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (150). The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (15) है, इसलिए \(q=10\times15=150\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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यदि \(\alpha,\beta\) समीकरण \(7x^2-25x+12=0\) के मूल हैं, तो \(\alpha\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(7x^2-25x+12=0\), what is \(\alpha\beta\)?
#quadratic
#product-of-roots
#formula
A \( \frac{12}{7}\)
B \( \frac{25}{7}\)
C (12)
D \(-\frac{12}{7}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{12}{7}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{12}{7}\). In exams, use \(\frac{c}{a}\) for the product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{12}{7}\). The product of roots is \(\frac{c}{a}=\frac{12}{7}\). In exams, use \(\frac{c}{a}\) for the product.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{12}{7}\) है। परीक्षा में गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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यदि \(9x^2-20x+p=0\) के मूलों का गुणनफल \(\frac{1}{3}\) है, तो (p) क्या होगा?
If the product of roots of \(9x^2-20x+p=0\) is \(\frac{1}{3}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (3)
B (9)
C \( \frac{1}{27}\)
D \( \frac{20}{3}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{9}\), so \(\frac{p}{9}=\frac{1}{3}\) gives (p=3). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (3). The product of roots is \(\frac{p}{9}\), so \(\frac{p}{9}=\frac{1}{3}\) gives (p=3). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{9}\) है, इसलिए \(\frac{p}{9}=\frac{1}{3}\) से (p=3) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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यदि \(x^2-23x+q=0\) का एक मूल (9) है, तो (q) का मान क्या होगा?
If one root of \(x^2-23x+q=0\) is (9), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (126)
B (207)
C (14)
D (23)
Explanation opens after your attempt
Step 1
Concept
The other root is (14), so \(q=9\times14=126\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (126). The other root is (14), so \(q=9\times14=126\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (14) है, इसलिए \(q=9\times14=126\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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यदि \(\alpha,\beta\) समीकरण \(5x^2-17x+6=0\) के मूल हैं, तो \(\alpha\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(5x^2-17x+6=0\), what is \(\alpha\beta\)?
#quadratic
#product-of-roots
#formula
A \( \frac{6}{5}\)
B \( \frac{17}{5}\)
C (6)
D \(-\frac{6}{5}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{6}{5}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{6}{5}\). In exams, use \(\frac{c}{a}\) for the product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{6}{5}\). The product of roots is \(\frac{c}{a}=\frac{6}{5}\). In exams, use \(\frac{c}{a}\) for the product.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{6}{5}\) है। परीक्षा में गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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यदि \(7x^2-15x+p=0\) के मूलों का गुणनफल \(\frac{2}{7}\) है, तो (p) क्या होगा?
If the product of roots of \(7x^2-15x+p=0\) is \(\frac{2}{7}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (2)
B (7)
C \( \frac{2}{49}\)
D \( \frac{15}{7}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{7}\), so \(\frac{p}{7}=\frac{2}{7}\) gives (p=2). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (2). The product of roots is \(\frac{p}{7}\), so \(\frac{p}{7}=\frac{2}{7}\) gives (p=2). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{7}\) है, इसलिए \(\frac{p}{7}=\frac{2}{7}\) से (p=2) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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यदि \(x^2-21x+q=0\) का एक मूल (8) है, तो (q) का मान क्या होगा?
If one root of \(x^2-21x+q=0\) is (8), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (104)
B (168)
C (13)
D (21)
Explanation opens after your attempt
Step 1
Concept
The other root is (13), so \(q=8\times13=104\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (104). The other root is (13), so \(q=8\times13=104\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (13) है, इसलिए \(q=8\times13=104\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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यदि \(\alpha,\beta\) समीकरण \(4x^2-13x+3=0\) के मूल हैं, तो \(\alpha\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(4x^2-13x+3=0\), what is \(\alpha\beta\)?
#quadratic
#product-of-roots
#formula
A \( \frac{3}{4}\)
B \( \frac{13}{4}\)
C (3)
D \(-\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{3}{4}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{3}{4}\). In exams, use \(\frac{c}{a}\) for the product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{3}{4}\). The product of roots is \(\frac{c}{a}=\frac{3}{4}\). In exams, use \(\frac{c}{a}\) for the product.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{3}{4}\) है। परीक्षा में गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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यदि \(6x^2-13x+p=0\) के मूलों का गुणनफल \(\frac{1}{2}\) है, तो (p) क्या होगा?
If the product of roots of \(6x^2-13x+p=0\) is \(\frac{1}{2}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (3)
B (6)
C \( \frac{1}{12}\)
D \( \frac{13}{2}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{6}\), so \(\frac{p}{6}=\frac{1}{2}\) gives (p=3). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (3). The product of roots is \(\frac{p}{6}\), so \(\frac{p}{6}=\frac{1}{2}\) gives (p=3). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{6}\) है, इसलिए \(\frac{p}{6}=\frac{1}{2}\) से (p=3) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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यदि \(x^2-15x+q=0\) का एक मूल (6) है, तो (q) का मान क्या होगा?
If one root of \(x^2-15x+q=0\) is (6), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (54)
B (90)
C (9)
D (15)
Explanation opens after your attempt
Step 1
Concept
The other root is (9), so \(q=6\times9=54\). In exams, when (a=1), (c) equals the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (54). The other root is (9), so \(q=6\times9=54\). In exams, when (a=1), (c) equals the product of roots.
Step 3
Exam Tip
दूसरा मूल (9) है, इसलिए \(q=6\times9=54\) होगा। परीक्षा में (a=1) हो तो (c) मूलों के गुणनफल के बराबर होता है।
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यदि \(\alpha,\beta\) समीकरण \(3x^2-11x+6=0\) के मूल हैं, तो \(\alpha\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(3x^2-11x+6=0\), what is \(\alpha\beta\)?
#quadratic
#product-of-roots
#formula
A (2)
B \( \frac{11}{3}\)
C (6)
D (-2)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{6}{3}=2\). In exams, use \(\frac{c}{a}\) for the product.
Step 2
Why this answer is correct
The correct answer is A. (2). The product of roots is \(\frac{c}{a}=\frac{6}{3}=2\). In exams, use \(\frac{c}{a}\) for the product.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{6}{3}=2\) है। परीक्षा में गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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यदि \(5x^2-11x+p=0\) के मूलों का गुणनफल \(\frac{2}{5}\) है, तो (p) क्या होगा?
If the product of roots of \(5x^2-11x+p=0\) is \(\frac{2}{5}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (2)
B (5)
C \( \frac{2}{25}\)
D \( \frac{11}{5}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{5}\), so \(\frac{p}{5}=\frac{2}{5}\) gives (p=2). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (2). The product of roots is \(\frac{p}{5}\), so \(\frac{p}{5}=\frac{2}{5}\) gives (p=2). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{5}\) है, इसलिए \(\frac{p}{5}=\frac{2}{5}\) से (p=2) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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यदि \(x^2-9x+q=0\) का एक मूल (4) है, तो (q) का मान क्या होगा?
If one root of \(x^2-9x+q=0\) is (4), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (20)
B (36)
C (5)
D (9)
Explanation opens after your attempt
Step 1
Concept
The other root is (5), so \(q=4\times5=20\). In exams, (c) equals the product of roots when (a=1).
Step 2
Why this answer is correct
The correct answer is A. (20). The other root is (5), so \(q=4\times5=20\). In exams, (c) equals the product of roots when (a=1).
Step 3
Exam Tip
दूसरा मूल (5) है, इसलिए \(q=4\times5=20\) होगा। परीक्षा में (c) मूलों के गुणनफल के बराबर होता है जब (a=1)।
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यदि \(\alpha,\beta\) समीकरण \(2x^2-9x+4=0\) के मूल हैं, तो \(\alpha\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(2x^2-9x+4=0\), what is \(\alpha\beta\)?
#quadratic
#product-of-roots
#formula
A (2)
B \( \frac{9}{2}\)
C (4)
D ( -2)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{4}{2}=2\). In exams, use \(\frac{c}{a}\) for the product.
Step 2
Why this answer is correct
The correct answer is A. (2). The product of roots is \(\frac{c}{a}=\frac{4}{2}=2\). In exams, use \(\frac{c}{a}\) for the product.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{4}{2}=2\) है। परीक्षा में गुणनफल के लिए \(\frac{c}{a}\) का प्रयोग करें।
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यदि \(4x^2-7x+p=0\) के मूलों का गुणनफल \(\frac{3}{2}\) है, तो (p) क्या होगा?
If the product of roots of \(4x^2-7x+p=0\) is \(\frac{3}{2}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (6)
B (3)
C \( \frac{3}{8}\)
D \( \frac{7}{2}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{4}\), so \(\frac{p}{4}=\frac{3}{2}\) gives (p=6). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (6). The product of roots is \(\frac{p}{4}\), so \(\frac{p}{4}=\frac{3}{2}\) gives (p=6). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{4}\) है, इसलिए \(\frac{p}{4}=\frac{3}{2}\) से (p=6) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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यदि \(x^2-5x+q=0\) का एक मूल (2) है, तो (q) का मान क्या होगा?
If one root of \(x^2-5x+q=0\) is (2), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (6)
B (10)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
The other root is (3), so \(q=2\times3=6\). In exams, connect (c) with product of roots.
Step 2
Why this answer is correct
The correct answer is A. (6). The other root is (3), so \(q=2\times3=6\). In exams, connect (c) with product of roots.
Step 3
Exam Tip
दूसरा मूल (3) है, इसलिए \(q=2\times3=6\) होगा। परीक्षा में (c) को मूलों के गुणनफल से जोड़ें।
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यदि \(x^2-7x+k=0\) की जड़ें एक-दूसरे की व्युत्क्रम हैं, तो (k) का मान क्या होगा?
If the roots of \(x^2-7x+k=0\) are reciprocals of each other, what is the value of (k)?
#quadratic-roots
#reciprocal-roots
#product-of-roots
A (1)
B (-1)
C (7)
D (49)
Explanation opens after your attempt
Step 1
Concept
For reciprocal roots, the product is (1), and here the product is (k). Hence (k=1); in exams, check the product first.
Step 2
Why this answer is correct
The correct answer is A. (1). For reciprocal roots, the product is (1), and here the product is (k). Hence (k=1); in exams, check the product first.
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए गुणनफल (1) होता है और यहाँ गुणनफल (k) है। इसलिए (k=1); परीक्षा में पहले गुणनफल देखें।
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यदि (3) और (-4) मूल हैं तो \(5x^2+px+q=0\) में \(\frac{q}{5}\) का मान क्या होगा?
If (3) and (-4) are roots, what is the value of \(\frac{q}{5}\) in \(5x^2+px+q=0\)?
#roots
#product_of_roots
#coefficient
A (-12)
B (12)
C (-1)
D (1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{q}{5}\). Since (3\cdot(-4)=-12), \(\frac{q}{5}=-12\).
Step 2
Why this answer is correct
The correct answer is A. (-12). The product of roots is \(\frac{q}{5}\). Since (3\cdot(-4)=-12), \(\frac{q}{5}=-12\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{q}{5}\) होता है। (3\cdot(-4)=-12) इसलिए \(\frac{q}{5}=-12\) है।
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यदि (3) और (5) समीकरण \(ax^2+bx+15=0\) के मूल हैं तो (a) का मान क्या है?
If (3) and (5) are roots of \(ax^2+bx+15=0\), what is the value of (a)?
#roots
#coefficient
#product_of_roots
A (1)
B (3)
C (5)
D (15)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{c}{a}\). From \(3\cdot5=\frac{15}{a}\), we get (a=1).
Step 2
Why this answer is correct
The correct answer is A. (1). The product of roots is \(\frac{c}{a}\). From \(3\cdot5=\frac{15}{a}\), we get (a=1).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}\) होता है। \(3\cdot5=\frac{15}{a}\) से (a=1) मिलता है।
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यदि (2) और (-3) मूल हैं तो \(3x^2+px+q=0\) में \(\frac{q}{3}\) का मान क्या होगा?
If (2) and (-3) are roots, what is the value of \(\frac{q}{3}\) in \(3x^2+px+q=0\)?
#roots
#product_of_roots
#coefficient
A (-6)
B (6)
C (-1)
D (1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{q}{3}\). Since (2\cdot(-3)=-6), \(\frac{q}{3}=-6\).
Step 2
Why this answer is correct
The correct answer is A. (-6). The product of roots is \(\frac{q}{3}\). Since (2\cdot(-3)=-6), \(\frac{q}{3}=-6\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{q}{3}\) होता है। (2\cdot(-3)=-6) इसलिए \(\frac{q}{3}=-6\) है।
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यदि \(x^2+6x+k=0\) के मूलों में एक दूसरे का वर्ग है और मूल (-2) तथा (-4) हैं तो (k) क्या होगा?
If the roots of \(x^2+6x+k=0\) are (-2) and (-4), where one is the square relation by value context, what is (k)?
#roots
#product_of_roots
#parameter
A (8)
B (6)
C (-8)
D (2)
Explanation opens after your attempt
Step 1
Concept
For roots (-2) and (-4), the product is (8). In a monic equation \(k=\alpha\beta=8\).
Step 2
Why this answer is correct
The correct answer is A. (8). For roots (-2) and (-4), the product is (8). In a monic equation \(k=\alpha\beta=8\).
Step 3
Exam Tip
मूल (-2) और (-4) होने पर गुणनफल (8) है। मोनिक समीकरण में \(k=\alpha\beta=8\) होता है।
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यदि (2) और (3) समीकरण \(ax^2+bx+6=0\) के मूल हैं तो (a) का मान क्या है?
If (2) and (3) are roots of \(ax^2+bx+6=0\), what is the value of (a)?
#roots
#coefficient
#product_of_roots
A (1)
B (2)
C (3)
D (6)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{c}{a}\). Here \(2\cdot3=\frac{6}{a}\), so (a=1).
Step 2
Why this answer is correct
The correct answer is A. (1). The product of roots is \(\frac{c}{a}\). Here \(2\cdot3=\frac{6}{a}\), so (a=1).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}\) होता है। यहां \(2\cdot3=\frac{6}{a}\) से (a=1) मिलता है।
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यदि \(\alpha\) और \(\beta\) समीकरण \(6x^2+5x-4=0\) के मूल हैं तो \(\alpha\beta\) क्या है?
If \(\alpha\) and \(\beta\) are roots of \(6x^2+5x-4=0\), what is \(\alpha\beta\)?
#roots
#product_of_roots
#calculation
A -\(\frac{2}{3}\)
B \(\frac{2}{3}\)
C -\(\frac{5}{6}\)
D \(\frac{5}{6}\)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{2}{3}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{-4}{6}=-\frac{2}{3}\). Do not forget the sign of (c).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{2}{3}\). The product of roots is \(\frac{c}{a}=\frac{-4}{6}=-\frac{2}{3}\). Do not forget the sign of (c).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{-4}{6}=-\frac{2}{3}\) होता है। (c) का चिन्ह न भूलें।
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समीकरण \(7x^2+4x-9=0\) के मूलों का गुणनफल क्या है?
What is the product of roots of \(7x^2+4x-9=0\)?
#roots
#product_of_roots
#formula
A -\(\frac{9}{7}\)
B \(\frac{9}{7}\)
C -\(\frac{4}{7}\)
D \(\frac{4}{7}\)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{9}{7}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}\). Here \(\frac{-9}{7}=-\frac{9}{7}\) is correct.
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{9}{7}\). The product of roots is \(\frac{c}{a}\). Here \(\frac{-9}{7}=-\frac{9}{7}\) is correct.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}\) होता है। यहां \(\frac{-9}{7}=-\frac{9}{7}\) सही है।
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यदि \(\alpha\) और \(\beta\) समीकरण \(5x^2+3x-2=0\) के मूल हैं तो \(\alpha\beta\) क्या है?
If \(\alpha\) and \(\beta\) are roots of \(5x^2+3x-2=0\), what is \(\alpha\beta\)?
#roots
#product_of_roots
#calculation
A -\(\frac{2}{5}\)
B \(\frac{2}{5}\)
C (3)
D (-3)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{2}{5}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{-2}{5}\). Do not forget the sign of (c).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{2}{5}\). The product of roots is \(\frac{c}{a}=\frac{-2}{5}\). Do not forget the sign of (c).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{-2}{5}\) होता है। (c) का चिन्ह न भूलें।
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समीकरण \(5x^2+6x-8=0\) के मूलों का गुणनफल क्या है?
What is the product of roots of \(5x^2+6x-8=0\)?
#roots
#product_of_roots
#formula
A -\(\frac{8}{5}\)
B \(\frac{8}{5}\)
C -\(\frac{6}{5}\)
D \(\frac{6}{5}\)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{8}{5}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}\). Here \(\frac{-8}{5}=-\frac{8}{5}\) is correct.
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{8}{5}\). The product of roots is \(\frac{c}{a}\). Here \(\frac{-8}{5}=-\frac{8}{5}\) is correct.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}\) होता है। यहां \(\frac{-8}{5}=-\frac{8}{5}\) सही है।
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यदि \(\alpha\) और \(\beta\) समीकरण \(4x^2+4x-3=0\) के मूल हैं तो \(\alpha\beta\) क्या है?
If \(\alpha\) and \(\beta\) are roots of \(4x^2+4x-3=0\), what is \(\alpha\beta\)?
#roots
#product_of_roots
#calculation
A -\(\frac{3}{4}\)
B \(\frac{3}{4}\)
C (1)
D (-1)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{3}{4}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}=\frac{-3}{4}\). Do not forget the sign of (c).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{3}{4}\). The product of roots is \(\frac{c}{a}=\frac{-3}{4}\). Do not forget the sign of (c).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}=\frac{-3}{4}\) होता है। (c) का चिन्ह न भूलें।
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समीकरण \(3x^2+8x-11=0\) के मूलों का गुणनफल क्या है?
What is the product of roots of \(3x^2+8x-11=0\)?
#roots
#product_of_roots
#formula
A \(-\frac{11}{3}\)
B \(\frac{11}{3}\)
C \(-\frac{8}{3}\)
D \(\frac{8}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{11}{3}\)
Step 1
Concept
The product of roots is \(\frac{c}{a}\). Here \(\frac{-11}{3}\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{11}{3}\). The product of roots is \(\frac{c}{a}\). Here \(\frac{-11}{3}\) is correct.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}\) होता है। यहां \(\frac{-11}{3}\) सही है।
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