Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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Showing 411-420 of 550 questions.
If the roots of \(x^2-2mx+36=0\) are equal, what are the possible values of (m)?
A. \(m=\pm6\)
समान मूलों के लिए (D=0), अतः ((-2m)2-144=0) होगा। इससे \(m^2=36\) और \(m=\pm6\) है। / For equal roots, (D=0), so ((-2m)2-144=0). This gives \(m^2=36\) and \(m=\pm6\).
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What is the condition for \(x^2+2px+25=0\) to have real roots?
A. \(p\leq -5\) या \(p\geq5\)\(p\leq -5\) or \(p\geq5\)
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(4p^2-100\geq0\), इसलिए \(p\leq-5\) या \(p\geq5\)। / For real roots, \(D\geq0\) is needed. Here \(4p^2-100\geq0\), so \(p\leq-5\) or \(p\geq5\).
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What is the sum of reciprocals of the roots of \(4x^2-13x+9=0\)?
A. \( \frac{13}{9} \)
व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{13}{4}}{\frac{9}{4}}=\frac{13}{9}\) है। / The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{13}{4}}{\frac{9}{4}}=\frac{13}{9}\).
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If (5) and (6) are roots of \(x^2-sx+p=0\), what is the value of (s+p)?
A. (41)
मूलों का योग (s=11) और गुणनफल (p=30) है। इसलिए (s+p=41) है। / The sum of roots gives (s=11) and the product gives (p=30). Therefore (s+p=41).
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If the roots of \(x^2-12x+k=0\) are real and distinct, what is the correct condition on (k)?
A. (k<36)
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (144-4k>0), इसलिए (k<36)। / For real and distinct roots, (D>0) is required. Here (144-4k>0), so (k<36).
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Which equation has discriminant (-23)?
A. \(x^2+x+6=0\)
\(x^2+x+6=0\) के लिए \(D=1^2-4\cdot1\cdot6=-23\) है। विवेचक में (4ac) पूरा घटाएं। / For \(x^2+x+6=0\), \(D=1^2-4\cdot1\cdot6=-23\). Subtract the full (4ac) in the discriminant.
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If one root of \(x^2+ax+24=0\) is (4), what are the other root and (a)?
A. दूसरा मूल (6), (a=-10)other root (6), (a=-10)
मूलों का गुणनफल (24) है, इसलिए दूसरा मूल (6) होगा। योग (10) है और (-a=10), इसलिए (a=-10)। / The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
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What is the sum of squares of the roots of \(5x^2+6x+1=0\)?
A. \( \frac{26}{25} \)
(\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) होता है। यहाँ (\left\(-\frac{6}{5}\right\)2-2\cdot\frac{1}{5}=\frac{26}{25})। / (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta). Here (\left\(-\frac{6}{5}\right\)2-2\cdot\frac{1}{5}=\frac{26}{25}).
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If \(\alpha\) and \(\beta\) are roots of \(x^2-9x+18=0\), what is (\(\alpha-3\)\(\beta-3\))?
A. (0)
(\(\alpha-3\)\(\beta-3\)=\alpha\beta-3\(\alpha+\beta\)+9) है। यहाँ (18-27+9=0)। / (\(\alpha-3\)\(\beta-3\)=\alpha\beta-3\(\alpha+\beta\)+9). Here (18-27+9=0).
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What is the standard form of ((x+3)(x+6)+(x-2)(x+5)=70)?
A. \(2x^2+12x-62=0\)
बाईं ओर \(x^2+9x+18+x^2+3x-10=2x^2+12x+8\) है। (70) घटाने पर \(2x^2+12x-62=0\) मिलता है। / The left side is \(x^2+9x+18+x^2+3x-10=2x^2+12x+8\). Subtracting (70) gives \(2x^2+12x-62=0\).
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