Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If (p(x)=x-2-2\sqrt{3}x+2), what is the product of its zeroes?
A. (2)
द्विघात \(ax^2+bx+c\) में शून्यकों का गुणनफल \(\frac{c}{a}\) होता है। यहाँ \(\frac{2}{1}=2\) है। / In \(ax^2+bx+c\), the product of zeroes is \(\frac{c}{a}\). Here it is \(\frac{2}{1}=2\).
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If (p(x)=x-2-10x+19), what are its zeroes?
A. \(5+\sqrt{6},5-\sqrt{6}\)
विविक्तकर (100-76=24) है, इसलिए शून्यक \(\frac{10\pm\sqrt{24}}{2}=5\pm\sqrt{6}\) हैं। परीक्षा में \(\sqrt{24}=2\sqrt{6}\) सरल करें। / The discriminant is (100-76=24), so the zeroes are \(\frac{10\pm\sqrt{24}}{2}=5\pm\sqrt{6}\). Simplify \(\sqrt{24}=2\sqrt{6}\) in exams.
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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 \(x^2-6x+1=0\), what type are the values of (x)?
C. दो अलग अपरिमेय वास्तविकTwo distinct irrational real values
विविक्तकर (36-4=32) है, जो पूर्ण वर्ग नहीं है और धनात्मक है। इसलिए दो अलग अपरिमेय वास्तविक मूल मिलते हैं। / The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.
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If zeroes of (p(x)=x-2-mx+9) are \(3\sqrt{2}\) and \(\frac{3}{\sqrt{2}}\), what is (m)?
A. \(\frac{9\sqrt{2}}{2}\)
योग \(3\sqrt{2}+\frac{3}{\sqrt{2}}=\frac{9\sqrt{2}}{2}\) है। इसलिए \(m=\frac{9\sqrt{2}}{2}\) होगा। / The sum is \(3\sqrt{2}+\frac{3}{\sqrt{2}}=\frac{9\sqrt{2}}{2}\). Hence \(m=\frac{9\sqrt{2}}{2}\).
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If \(x^2-4x+4=0\), what is the nature of the roots?
B. दो समान परिमेयTwo equal rational roots
विविक्तकर (16-16=0), इसलिए मूल समान हैं और (x=2) परिमेय है। परीक्षा में (D=0) का अर्थ समान मूल होता है। / The discriminant is (16-16=0), so the roots are equal and (x=2) is rational. In exams (D=0) means equal roots.
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If the sum of zeroes is \(2\sqrt{5}\) and the product is (4), which is a monic quadratic polynomial?
A. \(x^2-2\sqrt{5}x+4\)
\(मानक बहुपद (x^2-(\)योग)x+गुणनफल) होता है। इसलिए उत्तर \(x^2-2\sqrt{5}x+4\) है। \(/ A monic polynomial is (x^2-(\)sum)x+product\(). So the answer is (x^2-2\sqrt{5}x+4).\)
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If \(x^2+2x+5=0\), which statement about the roots in the real number system is correct?
C. कोई वास्तविक मूल नहीं हैThere are no real roots
विविक्तकर (4-20=-16) ऋणात्मक है, इसलिए वास्तविक मूल नहीं हैं। परीक्षा में ऋणात्मक विविक्तकर को वास्तविक शून्यक नहीं मानें। / The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.
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A. \(3\sqrt{2}\)
क्योंकि \(\sqrt{8}=2\sqrt{2}\), योग \(\sqrt{2}+2\sqrt{2}=3\sqrt{2}\) है। पहले करणी को सरल करें। / Since \(\sqrt{8}=2\sqrt{2}\), the sum is \(\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Simplify radicals first.
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If (p(x)=x-2-2ax+\(a^2-7\)) and (a) is rational, which statement about the zeroes is correct?
A. वे \(a+\sqrt{7}\) और \(a-\sqrt{7}\) हैंThey are \(a+\sqrt{7}\) and \(a-\sqrt{7}\)
(p(x)=(x-a)2-7), इसलिए \(x=a\pm\sqrt{7}\) है। पूर्ण वर्ग रूप पहचानना कठिन प्रश्नों में समय बचाता है। / (p(x)=(x-a)2-7), so \(x=a\pm\sqrt{7}\). Recognizing a perfect-square form saves time in hard questions.
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