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1536 questions tagged with hard.
If both roots of \(x^2+bx+25=0\) are equal and their value is (-5), what is (b)?
A. (10)
दोनों मूल (-5) हैं, इसलिए योग (-10) है। यहाँ मूलों का योग (-b) है, अतः (b=10)। / Both roots are (-5), so their sum is (-10). Here the sum of roots is (-b), so (b=10).
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If \(x^2+6x+c=0\) is a perfect square quadratic equation, what is the value of (c)?
A. (9)
पूर्ण वर्ग के लिए (x-2+6x+c=(x+3)2) होना चाहिए। इसलिए (c=9) होगा। / For a perfect square, (x-2+6x+c=(x+3)2) is needed. Hence (c=9).
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What is the difference between the roots of \(3x^2-10x+8=0\)?
A. \( \frac{2}{3} \)
मूल (2) और \(\frac{4}{3}\) हैं। उनका अंतर \(2-\frac{4}{3}=\frac{2}{3}\) है। / The roots are (2) and \(\frac{4}{3}\). Their difference is \(2-\frac{4}{3}=\frac{2}{3}\).
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If one root of (x-2-(m+2)x+2m=0) is (2), what is the other root?
A. (m)
गुणनफल (2m) है और एक मूल (2) है। इसलिए दूसरा मूल \(\frac{2m}{2}=m\) है। / The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
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Which statement is correct for \(x^2-4x+6=0\)?
A. वास्तविक मूल नहीं हैंIt has no real roots
यहाँ (D=(-4)2-4\cdot1\cdot6=-8<0) है। इसलिए वास्तविक मूल नहीं होंगे। / Here (D=(-4)2-4\cdot1\cdot6=-8<0). Therefore it has no real roots.
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If the roots of \(x^2+px+q=0\) are (p) and (q) themselves, which relation is correct?
A. (p+q=-p) और (pq=q)(p+q=-p) and (pq=q)
मूलों का योग (-p) और गुणनफल (q) होता है। दिए मूल (p,q) हैं, इसलिए (p+q=-p) और (pq=q)। / The sum of roots is (-p) and the product is (q). Since the roots are (p,q), we get (p+q=-p) and (pq=q).
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What is the standard form of ((x+1)(x+2)+(x+3)(x+4)=50)?
A. \(2x^2+10x-36=0\)
बाईं ओर \(x^2+3x+2+x^2+7x+12=2x^2+10x+14\) है। (50) घटाने पर \(2x^2+10x-36=0\) मिलता है। / The left side is \(x^2+3x+2+x^2+7x+12=2x^2+10x+14\). Subtracting (50) gives \(2x^2+10x-36=0\).
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In which option will the given equation not remain quadratic?
A. ((t-2)x-2+5x+1=0), (t=2)
पहले विकल्प में (t=2) रखने पर \(x^2\) का गुणांक (0) हो जाता है। तब समीकरण रैखिक बन जाता है। / In the first option, putting (t=2) makes the coefficient of \(x^2\) equal to (0). Then the equation becomes linear.
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If \(\alpha\) and \(\beta\) are roots of \(x^2-8x+15=0\), what is (\(\alpha-1\)\(\beta-1\))?
A. (8)
(\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1) होता है। यहाँ (15-8+1=8)। / (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1). Here (15-8+1=8).
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What is the sum of squares of the roots of \(2x^2+7x+3=0\)?
A. \( \frac{37}{4} \)
यदि मूल \(\alpha,\beta\) हैं, तो (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta)। यहाँ (\left\(-\frac{7}{2}\right\)2-2\cdot\frac{3}{2}=\frac{37}{4})। / If the roots are \(\alpha,\beta\), then (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta). Here (\left\(-\frac{7}{2}\right\)2-2\cdot\frac{3}{2}=\frac{37}{4}).
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If one root of \(x^2+ax+12=0\) is (3), what are the other root and (a)?
A. दूसरा मूल (4), (a=-7)other root (4), (a=-7)
मूलों का गुणनफल (12) है, इसलिए दूसरा मूल (4) होगा। योग (7) है और (-a=7), इसलिए (a=-7)। / The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
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Which equation has discriminant (-15)?
A. \(x^2+x+4=0\)
\(x^2+x+4=0\) के लिए \(D=1^2-4\cdot1\cdot4=-15\) है। विवेचक निकालते समय (4ac) पूरा घटाएं। / For \(x^2+x+4=0\), \(D=1^2-4\cdot1\cdot4=-15\). Subtract the full (4ac) while finding the discriminant.
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If the roots of \(x^2-6x+k=0\) are real and distinct, what is the correct condition on (k)?
A. (k<9)
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (36-4k>0), इसलिए (k<9)। / For real and distinct roots, (D>0) is needed. Here (36-4k>0), so (k<9).
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If (2) and (3) are roots of \(x^2-sx+p=0\), what is the value of (s+p)?
A. (11)
मूलों का योग (s=5) और गुणनफल (p=6) है। इसलिए (s+p=11) है। / The sum of roots gives (s=5) and the product gives (p=6). Therefore (s+p=11).
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What is the sum of reciprocals of the roots of \(2x^2-5x+3=0\)?
A. \( \frac{5}{3} \)
व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\) है। / The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\).
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What is the condition for \(x^2+2kx+9=0\) to have real roots?
A. \(k\leq -3\) या \(k\geq 3\)\(k\leq -3\) or \(k\geq 3\)
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(4k^2-36\geq0\), इसलिए \(k\leq-3\) या \(k\geq3\)। / For real roots, \(D\geq0\) is needed. Here \(4k^2-36\geq0\), so \(k\leq-3\) or \(k\geq3\).
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If the roots of \(x^2-2kx+16=0\) are equal, what are the possible values of (k)?
A. \(k=\pm4\)
समान मूलों के लिए (D=0), इसलिए ((-2k)2-64=0) मिलता है। इससे \(k^2=16\) और \(k=\pm4\) है। / For equal roots, (D=0), so ((-2k)2-64=0). This gives \(k^2=16\) and \(k=\pm4\).
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Which monic quadratic equation has sum of roots (7) and product (10)?
A. \(x^2-7x+10=0\)
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। इसलिए \(x^2-7x+10=0\) सही है। \(/ A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
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If (x=-2) is a root of \(3x^2+px+10=0\), what is the value of (p)?
A. (11)
(x=-2) रखने पर (12-2p+10=0) मिलता है। इससे (p=11) है। / Putting (x=-2) gives (12-2p+10=0). Hence (p=11).
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Write \(\frac{x^2-3}{2}+\frac{x-1}{3}=4\) in standard form with integer coefficients.
A. \(3x^2+2x-29=0\)
पूरे समीकरण को (6) से गुणा करने पर \(3x^2-9+2x-2=24\) मिलता है। इसलिए \(3x^2+2x-35=0\) नहीं बल्कि सही रूप \(3x^2+2x-35=0\) होगा। / Multiplying the whole equation by (6) gives \(3x^2-9+2x-2=24\). Thus the standard form is \(3x^2+2x-35=0\).
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What is the standard form of ((x-2)2+(2x+1)2=25)?
D. \(5x^2-20=0\)
विस्तार करने पर \(x^2-4x+4+4x^2+4x+1=25\) मिलता है। इससे \(5x^2-20=0\) बनता है। / Expanding gives \(x^2-4x+4+4x^2+4x+1=25\). This gives \(5x^2-20=0\).
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If (\(m^2-9\)x-2+4x-5=0) is a quadratic equation, what is the correct condition on (m)?
C. \(m\neq \pm3\)
द्विघात होने के लिए \(m^2-9\neq0\) होना चाहिए। इसलिए \(m\neq3\) और \(m\neq-3\) दोनों जरूरी हैं। / For the equation to be quadratic, \(m^2-9\neq0\) is needed. Hence both \(m\neq3\) and \(m\neq-3\) are necessary.
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What is the standard quadratic form of ((5x-2)(2x+3)=7x)?
A. \(10x^2+4x-6=0\)
((5x-2)(2x+3)=10x-2+11x-6) है और (7x) घटाने पर \(10x^2+4x-6=0\) मिलता है। पहले विस्तार करें फिर सभी पद एक ओर लाएं। / Here ((5x-2)(2x+3)=10x-2+11x-6) and subtracting (7x) gives \(10x^2+4x-6=0\). First expand and then bring all terms to one side.
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If \(x=\frac{1}{\sqrt{5}-2}\), what is the simplified form of (x)?
A. \(\sqrt{5}+2\)
\(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\) है। परीक्षा में हर का परिमेयकरण करें। / \(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\). Rationalise the denominator in exams.
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Which polynomial has rational coefficients and zeroes \(1+\sqrt{2}\) and \(1-\sqrt{2}\)?
A. \(x^2-2x-1\)
योग (2) और गुणनफल (1-2=-1), इसलिए बहुपद \(x^2-2x-1\) है। परीक्षा में चिन्हों को ठीक रखें। / The sum is (2) and the product is (1-2=-1), so the polynomial is \(x^2-2x-1\). Keep signs correct in exams.
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If (p(x)=x-2+6x+6), what is the correct classification of its zeroes?
A. वास्तविक और अपरिमेयReal and irrational
(D=36-24=12) है, जो धनात्मक है पर पूर्ण वर्ग नहीं है। इसलिए शून्यक वास्तविक और अपरिमेय हैं। / (D=36-24=12), which is positive but not a perfect square. Hence the zeroes are real and irrational.
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A. \(\frac{10}{19}\)
शून्यकों का योग (10) और गुणनफल (19) है। अतः \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{10}{19}\)। / The sum of zeroes is (10) and the product is (19). Hence \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{10}{19}\).
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If (p(x)=2x-2-4\sqrt{2}x+4), what is the sum of its zeroes?
A. \(2\sqrt{2}\)
योग \(-\frac{b}{a}=-\frac{-4\sqrt{2}}{2}=2\sqrt{2}\) है। गुणांक देखते समय (a) को मत भूलें। / The sum is \(-\frac{b}{a}=-\frac{-4\sqrt{2}}{2}=2\sqrt{2}\). Do not forget the coefficient (a).
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If (p(x)=x-2-8x+10), what are its zeroes?
A. \(4+\sqrt{6}\) और \(4-\sqrt{6}\)\(4+\sqrt{6}\) and \(4-\sqrt{6}\)
सूत्र से शून्यक \(\frac{8\pm\sqrt{64-40}}{2}=4\pm\sqrt{6}\) हैं। पहले विविक्तकर सरल करें। / By the formula, the zeroes are \(\frac{8\pm\sqrt{64-40}}{2}=4\pm\sqrt{6}\). Simplify the discriminant first.
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After how many decimal places will \(\frac{2^5\cdot 5^2}{2^{10}\cdot 5^6}\) terminate?
B. (5)
कटौती के बाद हर \(2^5\cdot 5^4\) बचेगा। बड़ी घात (5) है इसलिए दशमलव (5) स्थानों पर समाप्त होगा। / After cancellation, the denominator becomes \(2^5\cdot 5^4\). The larger exponent is (5), so the decimal terminates after (5) places.
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