roots se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
625 questions tagged with roots.
What are the roots of \(x^2-16x=0\)?
A. (x=0,16)
(x(x-16)=0), इसलिए (x=0) या (x=16) है। परीक्षा में शून्य गुणनफल नियम से हर गुणनखंड को अलग रखें। / (x(x-16)=0), so (x=0) or (x=16). In exams, use zero product rule on each factor separately.
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What are the roots of \(x^2+x-6=0\)?
A. (x=2,-3)
((x+3)(x-2)=0), इसलिए (x=-3) और (x=2) हैं। परीक्षा में गुणनखंड से मूल निकालते समय संकेत बदलता है। / ((x+3)(x-2)=0), so (x=-3) and (x=2). In exams, the sign changes while finding roots from factors.
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Solving ((x-2)2=9) gives which values?
A. (x=5,-1)
\(x-2=\pm3\), इसलिए (x=5) या (x=-1) है। परीक्षा में \(\pm\) के दोनों केस लिखें। / \(x-2=\pm3\), so (x=5) or (x=-1). In exams, write both \(\pm\) cases.
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Which are the correct roots for \(x^2+6x+8=0\)?
A. (x=-2,-4)
(x-2+6x+8=(x+2)(x+4)), इसलिए (x=-2) और (x=-4) हैं। परीक्षा में धनात्मक मध्य पद के साथ ऋणात्मक मूल आ सकते हैं। / (x-2+6x+8=(x+2)(x+4)), so (x=-2) and (x=-4). In exams, a positive middle term can give negative roots.
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What are the roots of \(x^2+3x+2=0\)?
A. (x=-1,-2)
((x+1)(x+2)=0), इसलिए (x=-1) और (x=-2) हैं। परीक्षा में ((x+a)=0) से (x=-a) लिखें। / ((x+1)(x+2)=0), so (x=-1) and (x=-2). In exams, from ((x+a)=0), write (x=-a).
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What are the solutions of \(x^2-25=0\)?
A. \(x=\pm5\)
(x-2-25=(x-5)(x+5)), इसलिए \(x=\pm5\) है। परीक्षा में \(25=5^2\) पहचानें। / (x-2-25=(x-5)(x+5)), so \(x=\pm5\). In exams, recognize \(25=5^2\).
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What are the roots of \(x^2-7x+10=0\)?
A. (x=2,5)
(x-2-7x+10=(x-2)(x-5)), इसलिए मूल (2) और (5) हैं। परीक्षा में (2+5=7) और \(2\times5=10\) जांचें। / (x-2-7x+10=(x-2)(x-5)), so the roots are (2) and (5). In exams, check (2+5=7) and \(2\times5=10\).
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Which is the correct formula to find roots of \(ax^2+bx+c=0\) by quadratic formula?
A. \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
द्विघात सूत्र \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) है। परीक्षा में (a), (b), (c) को सही पहचानना सबसे जरूरी है। / The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). In exams, identifying (a), (b), and (c) correctly is most important.
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If \(x^2=16\), what is the value of (x) by square root method?
A. \(x=\pm4\)
\(x^2=16\) से \(x=\pm\sqrt{16}=\pm4\) मिलता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें। / From \(x^2=16\), \(x=\pm\sqrt{16}=\pm4\). In exams, do not forget \(\pm\) while taking square root.
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Using factorisation method, what are the roots of \(x^2-5x+6=0\)?
A. (x=2,3)
(x-2-5x+6=(x-2)(x-3)), इसलिए मूल (2) और (3) हैं। परीक्षा में पहले ऐसे दो संख्याएं खोजें जिनका गुणनफल (6) और योग (-5) हो। / (x-2-5x+6=(x-2)(x-3)), so the roots are (2) and (3). In exams, first find two numbers whose product is (6) and sum is (-5).
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A. \(\frac{13}{36}\)
यहां \(\alpha+\beta=11\) और \(\alpha\beta=24\) है। मान \(\frac{\alpha+\beta+2}{\alpha\beta+\alpha+\beta+1}=\frac{13}{36}\) होगा। / Here \(\alpha+\beta=11\) and \(\alpha\beta=24\). The value is \(\frac{\alpha+\beta+2}{\alpha\beta+\alpha+\beta+1}=\frac{13}{36}\).
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A. (-24)
यहां \(\alpha+\beta=3\) और \(\alpha\beta=-28\) है। (\(\alpha+1\)\(\beta+1\)=-28+3+1=-24) है। / Here \(\alpha+\beta=3\) and \(\alpha\beta=-28\). (\(\alpha+1\)\(\beta+1\)=-28+3+1=-24).
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A. (20)
यहां \(\alpha+\beta=5\) और \(\alpha\beta=5\) है। (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=20) है। / Here \(\alpha+\beta=5\) and \(\alpha\beta=5\). (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=20).
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A. \(x^2-\frac{7}{10}x+\frac{1}{10}=0\)
पुराने योग (7) और गुणनफल (10) हैं। उलटे मूलों का योग \(\frac{7}{10}\) और गुणनफल \(\frac{1}{10}\) होगा। / The old sum is (7) and product is (10). The reciprocal roots have sum \(\frac{7}{10}\) and product \(\frac{1}{10}\).
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If roots of \(3x^2+px+18=0\) are (-2) and (-3), what is the value of (p)?
A. (15)
मूलों का योग (-5) है और \(-\frac{p}{3}=-5\) होगा। इसलिए (p=15) है। / The sum of roots is (-5), and \(-\frac{p}{3}=-5\). Therefore (p=15).
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A. \(\frac{106}{45}\)
यहां \(\alpha+\beta=14\) और \(\alpha\beta=45\) है। \(\alpha^2+\beta^2=196-90=106\) इसलिए मान \(\frac{106}{45}\) है। / Here \(\alpha+\beta=14\) and \(\alpha\beta=45\). \(\alpha^2+\beta^2=196-90=106\), so the value is \(\frac{106}{45}\).
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If roots of \(x^2+kx+49=0\) are equal and negative, what is (k)?
A. (14)
बराबर ऋणात्मक मूल (-7) और (-7) हैं। योग (-14) है इसलिए (k=14) है। / The equal negative roots are (-7) and (-7). Their sum is (-14), so (k=14).
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A. (7)
यहां \(\alpha+\beta=4\) और \(\alpha\beta=-5\) है। (\(\alpha+2\)\(\beta+2\)=-5+2(4)+4=7) है। / Here \(\alpha+\beta=4\) and \(\alpha\beta=-5\). (\(\alpha+2\)\(\beta+2\)=-5+2(4)+4=7).
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A. (9) या (-9)(9) or (-9)
मूल (5) और (4) हैं इसलिए योग (9) और अंतर \(\pm1\) हो सकता है। इसलिए मान (9) या (-9) है। / The roots are (5) and (4), so the sum is (9) and the difference can be \(\pm1\). Hence the value is (9) or (-9).
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If \(\alpha\) and \(\beta\) are roots of \(x^2-8x+12=0\), what is the value of \(\alpha^2\beta^2\)?
A. (144)
(\alpha-2\beta-2=\(\alpha\beta\)2) होता है। यहां \(\alpha\beta=12\) इसलिए मान (144) है। / (\alpha-2\beta-2=\(\alpha\beta\)2). Here \(\alpha\beta=12\), so the value is (144).
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A. (1)
समीकरण के मूल (6) और (7) हैं। इसलिए छोटा मूल (6) और बड़ा मूल (7) है तथा \(\beta-\alpha=1\) है। / The roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so \(\beta-\alpha=1\).
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A. (8)
यहां \(\alpha+\beta=10\) और \(\alpha\beta=24\) है। (\(\alpha-2\)\(\beta-2\)=24-20+4=8) है। / Here \(\alpha+\beta=10\) and \(\alpha\beta=24\). (\(\alpha-2\)\(\beta-2\)=24-20+4=8).
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Which can be the roots of (x-2-(2m+3)x+(m+1)(m+2)=0)?
A. (m+1) और (m+2)(m+1) and (m+2)
(m+1+m+2=2m+3) और गुणनफल ((m+1)(m+2)) है। इसलिए मूल (m+1) और (m+2) हो सकते हैं। / (m+1+m+2=2m+3), and the product is ((m+1)(m+2)). Therefore the roots can be (m+1) and (m+2).
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A. (6)
यहां \(\alpha+\beta=-2\) और \(\alpha\beta=-2\) है। (\alpha-2+\alpha\beta+\beta-2=\(\alpha+\beta\)2-\alpha\beta=6) है। / Here \(\alpha+\beta=-2\) and \(\alpha\beta=-2\). (\alpha-2+\alpha\beta+\beta-2=\(\alpha+\beta\)2-\alpha\beta=6).
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If \(\alpha\) and \(\beta\) are roots of \(x^2-6x+2=0\), what is the value of \(\alpha^2+\beta^2\)?
A. (32)
यहां \(\alpha+\beta=6\) और \(\alpha\beta=2\) है। \(\alpha^2+\beta^2=36-4=32\) है। / Here \(\alpha+\beta=6\) and \(\alpha\beta=2\). \(\alpha^2+\beta^2=36-4=32\).
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If roots of \(x^2+px+q=0\) are (5) and (-2), what is the value of (p-q)?
A. (7)
योग (3) है इसलिए (p=-3) और गुणनफल (-10) है इसलिए (q=-10)। अतः (p-q=-3-(-10)=7) है। / The sum is (3), so (p=-3), and the product is (-10), so (q=-10). Thus (p-q=-3-(-10)=7).
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If (3) and (-4) are roots, what is the value of \(\frac{q}{5}\) in \(5x^2+px+q=0\)?
A. (-12)
मूलों का गुणनफल \(\frac{q}{5}\) होता है। (3\cdot(-4)=-12) इसलिए \(\frac{q}{5}=-12\) है। / The product of roots is \(\frac{q}{5}\). Since (3\cdot(-4)=-12), \(\frac{q}{5}=-12\).
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A. (4)
यहां \(\alpha+\beta=9\) और \(\alpha\beta=18\) है। (\(\alpha-2\)\(\beta-2\)=18-2(9)+4=4) है। / Here \(\alpha+\beta=9\) and \(\alpha\beta=18\). (\(\alpha-2\)\(\beta-2\)=18-2(9)+4=4).
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If roots of (x-2-2(m+2)x+(m+2)2=0) are equal, what is true for (m)?
A. हर वास्तविक (m)Every real (m)
समीकरण ((x-(m+2))2=0) बनता है। इसलिए हर वास्तविक (m) के लिए दोनों मूल बराबर होंगे। / The equation becomes ((x-(m+2))2=0). Therefore the roots are equal for every real (m).
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If (x-2-4x+(m+4)=0) has no real roots, which condition on (m) is correct?
A. (m>0)
वास्तविक मूल नहीं होने के लिए (D<0) चाहिए। (16-4(m+4)<0) से (m>0) मिलता है। / For no real roots (D<0) is required. From (16-4(m+4)<0), we get (m>0).
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