roots se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
632 questions tagged with roots.
What is the repeated root of \(16x^2-8x+1=0\)?
A. \(\frac{1}{4}\)
(16x-2-8x+1=(4x-1)2) है। इसलिए दोहराया मूल \(\frac{1}{4}\) है। / (16x-2-8x+1=(4x-1)2). Therefore the repeated root is \(\frac{1}{4}\).
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If the sum of roots is (0) and product is (-25), which monic equation is formed?
A. \(x^2-25=0\)
मोनिक समीकरण (x-2-(0)x+(-25)=0) होगा। इसलिए \(x^2-25=0\) सही है। / The monic equation is (x-2-(0)x+(-25)=0). Therefore \(x^2-25=0\) is correct.
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A. (23)
यहां \(\alpha+\beta=8\) और \(\alpha\beta=15\) है। इसलिए कुल (8+15=23) है। / Here \(\alpha+\beta=8\) and \(\alpha\beta=15\). Therefore the total is (8+15=23).
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What are the roots of \(x^2+x-20=0\)?
A. (4) और (-5)(4) and (-5)
(x-2+x-20=(x+5)(x-4)) है। इसलिए मूल (-5) और (4) हैं। / (x-2+x-20=(x+5)(x-4)). Therefore the roots are (-5) and (4).
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If the product of roots of a quadratic equation is negative, which statement is correct?
A. एक मूल धनात्मक और दूसरा ऋणात्मक हैOne root is positive and the other is negative
ऋणात्मक गुणनफल तभी होता है जब मूलों के चिन्ह विपरीत हों। इसलिए एक मूल धनात्मक और दूसरा ऋणात्मक होगा। / A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.
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If a quadratic equation has (a=1), (b=-16), and (c=64), what will be the roots?
A. (8) और (8)(8) and (8)
समीकरण \(x^2-16x+64=0\) है जो ((x-8)2=0) बनता है। इसलिए दोनों मूल (8) हैं। / The equation is \(x^2-16x+64=0\), which becomes ((x-8)2=0). Therefore both roots are (8).
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What is the nature of roots of \(9x^2-24x+16=0\)?
A. बराबर वास्तविक मूलEqual real roots
यह (9x-2-24x+16=(3x-4)2) है। इसलिए दोनों मूल \(\frac{4}{3}\) और \(\frac{4}{3}\) बराबर हैं। / It is (9x-2-24x+16=(3x-4)2). Therefore both roots are \(\frac{4}{3}\) and \(\frac{4}{3}\), which are equal.
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If \(x=\frac{4}{3}\) is a root of \(3x^2-7x+c=0\), what is the value of (c)?
A. (4)
\(x=\frac{4}{3}\) रखने पर \(\frac{16}{3}-\frac{28}{3}+c=0\) इसलिए (c=4) है। भिन्न मूल में हर पद अलग निकालें। / Putting \(x=\frac{4}{3}\) gives \(\frac{16}{3}-\frac{28}{3}+c=0\), so (c=4). For fractional roots calculate each term separately.
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If both roots of \(x^2+px+25=0\) are (-5) and (-5), what is the value of (p)?
A. (10)
मूलों का योग (-10) है और (-p=-10) होगा। इसलिए (p=10) है। / The sum of roots is (-10), and (-p=-10). Therefore (p=10).
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If (x-4) and (x+7) are factors of a quadratic equation, what are its roots?
A. (4) और (-7)(4) and (-7)
(x-4=0) से (x=4) और (x+7=0) से (x=-7) मिलता है। गुणनखंड का चिन्ह उलटकर मूल मिलता है। / From (x-4=0), (x=4), and from (x+7=0), (x=-7). The sign changes when finding a root from a factor.
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For \(3x^2-14x+8=0\), what are the sum and product of roots respectively?
A. \(\frac{14}{3}\) और \(\frac{8}{3}\)\(\frac{14}{3}\) and \(\frac{8}{3}\)
योग \(-\frac{b}{a}=\frac{14}{3}\) और गुणनफल \(\frac{c}{a}=\frac{8}{3}\) है। पहले (a), (b), (c) पहचानें। / The sum is \(-\frac{b}{a}=\frac{14}{3}\) and the product is \(\frac{c}{a}=\frac{8}{3}\). Identify (a), (b), and (c) first.
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A. \(\frac{5}{14}\)
यहां \(\alpha+\beta=-5\) और \(\alpha\beta=-14\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\) है। / Here \(\alpha+\beta=-5\) and \(\alpha\beta=-14\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\).
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A. (25)
(\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta=49-24=25) है। ऐसे प्रश्नों में पहचान का प्रयोग करें। / (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta=49-24=25). Use identities in such questions.
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A. (12)
नए मूलों का योग (3\alpha+3\beta=3\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है। / The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
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If one root of \(x^2-15x+54=0\) is (6), what is the other root?
A. (9)
मूलों का गुणनफल (54) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{54}{6}=9\) है। / The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
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If the sum of roots is (9) and product is (18), which monic quadratic equation is formed?
A. \(x^2-9x+18=0\)
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। इसलिए \(x^2-9x+18=0\) सही है। \(/ A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-9x+18=0) is correct.\)
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For \(x^2-8x+k=0\) to have equal real roots, what should be the value of (k)?
A. (16)
बराबर मूलों के लिए (D=0) होता है। (64-4k=0) से (k=16) मिलता है। / For equal roots (D=0). From (64-4k=0), we get (k=16).
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Which statement is correct about the real roots of \(x^2-4x+13=0\)?
C. कोई वास्तविक मूल नहीं हैIt has no real roots
यहां (D=(-4)2-4(1)(13)=-36<0) है। इसलिए कोई वास्तविक मूल नहीं है। / Here (D=(-4)2-4(1)(13)=-36<0). Therefore there are no real roots.
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If (D=64) for a quadratic equation, what will be the nature of its real roots?
A. दो भिन्न वास्तविक मूलTwo distinct real roots
क्योंकि (D=64>0) है इसलिए दो भिन्न वास्तविक मूल होंगे। (D>0) होने पर मूल अलग अलग होते हैं। / Since (D=64>0), there will be two distinct real roots. When (D>0), the roots are different.
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What is the discriminant (D) of \(3x^2-6x+3=0\)?
A. (0)
यहां \(D=b^2-4ac=36-36=0\) है। इसलिए दोनों वास्तविक मूल बराबर होंगे। / Here \(D=b^2-4ac=36-36=0\). Therefore the two real roots will be equal.
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What is the product of roots of \(5x^2+6x-8=0\)?
A. -\(\frac{8}{5}\)
मूलों का गुणनफल \(\frac{c}{a}\) होता है। यहां \(\frac{-8}{5}=-\frac{8}{5}\) सही है। / The product of roots is \(\frac{c}{a}\). Here \(\frac{-8}{5}=-\frac{8}{5}\) is correct.
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What is the sum of roots of \(4x^2-11x+7=0\)?
A. \(\frac{11}{4}\)
मूलों का योग \(-\frac{b}{a}\) होता है। यहां (a=4) और (b=-11) इसलिए योग \(\frac{11}{4}\) है। / The sum of roots is \(-\frac{b}{a}\). Here (a=4) and (b=-11), so the sum is \(\frac{11}{4}\).
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If (3) is a root of \(x^2+kx-18=0\), what is the value of (k)?
A. (3)
(x=3) रखने पर (9+3k-18=0) इसलिए (k=3) है। पैरामीटर वाले प्रश्न में मूल को सीधे रखें। / Putting (x=3) gives (9+3k-18=0), so (k=3). In parameter questions substitute the root directly.
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Which equation has roots (-3) and (6)?
A. \(x^2-3x-18=0\)
मूल (-3) और (6) होने पर ((x+3)(x-6)=0) होगा। इसे खोलने पर \(x^2-3x-18=0\) मिलता है। / With roots (-3) and (6), we get ((x+3)(x-6)=0). Expanding it gives \(x^2-3x-18=0\).
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What are the roots of \(2x^2+9x+4=0\)?
A. (-4) और \(-\frac{1}{2}\)(-4) and \(-\frac{1}{2}\)
(2x-2+9x+4=(2x+1)(x+4)) है। इसलिए मूल \(-\frac{1}{2}\) और (-4) हैं। / (2x-2+9x+4=(2x+1)(x+4)). Therefore the roots are \(-\frac{1}{2}\) and (-4).
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What are the roots of \(3x^2-13x+4=0\)?
A. (4) और \(\frac{1}{3}\)(4) and \(\frac{1}{3}\)
(3x-2-13x+4=(3x-1)(x-4)) है। इसलिए मूल \(\frac{1}{3}\) और (4) हैं। / (3x-2-13x+4=(3x-1)(x-4)). Therefore the roots are \(\frac{1}{3}\) and (4).
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What are the roots of \(x^2-12x+35=0\)?
A. (5) और (7)(5) and (7)
(x-2-12x+35=(x-5)(x-7)) है। इसलिए मूल (5) और (7) हैं। / (x-2-12x+35=(x-5)(x-7)). Therefore the roots are (5) and (7).
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Is (x=-2) a root of \(3x^2+2x-8=0\)?
A. हाँYes
(x=-2) रखने पर (12-4-8=0) मिलता है। ऋणात्मक मूल रखते समय चिन्हों को ध्यान से संभालें। / Putting (x=-2) gives (12-4-8=0). Handle signs carefully when substituting a negative root.
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Is (x=4) a root of \(x^2-9x+20=0\)?
A. हाँYes
(x=4) रखने पर (16-36+20=0) मिलता है। इसलिए (4) इस समीकरण का मूल है। / Putting (x=4) gives (16-36+20=0). Therefore (4) is a root of this equation.
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If substituting (x=a) in a quadratic equation (p(x)=0) gives (p(a)=0), what is (a) called?
A. मूलRoot
क्योंकि (x=a) रखने पर समीकरण सत्य हो जाता है इसलिए (a) मूल है। परीक्षा में मूल की जांच सीधे प्रतिस्थापन से करें। / Since the equation becomes true after substituting (x=a), (a) is a root. In exams check a root by direct substitution.
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