Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 241-250 of 600 questions.
Solving ((x+6)2=5), what will be the value of (x)?
A. \(x=-6\pm\sqrt{5}\)
\(x+6=\pm\sqrt{5}\), इसलिए \(x=-6\pm\sqrt{5}\) है। परीक्षा में \(\pm\) को पूरे वर्गमूल के साथ लिखें। / \(x+6=\pm\sqrt{5}\), so \(x=-6\pm\sqrt{5}\). In exams, write \(\pm\) with the whole square root.
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Which is the correct solution for \(x^2+12x+32=0\)?
A. (x=-4,-8)
(x-2+12x+32=(x+4)(x+8)), इसलिए (x=-4,-8) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद से ऋणात्मक मूल आ सकते हैं। / (x-2+12x+32=(x+4)(x+8)), so (x=-4,-8). In exams, a positive middle term and positive constant can give negative roots.
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Which factorised form is correct for \(7x^2+8x+1=0\)?
A. ((7x+1)(x+1)=0)
((7x+1)(x+1)=7x-2+8x+1), इसलिए यह सही है। परीक्षा में गुणनखंड को विस्तार करके जांचें। / ((7x+1)(x+1)=7x-2+8x+1), so it is correct. In exams, verify the factors by expanding.
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What will be the roots of \(7x^2+8x+1=0\)?
A. \(x=-\frac{1}{7},-1\)
((7x+1)(x+1)=0), इसलिए \(x=-\frac{1}{7}\) और (-1) हैं। परीक्षा में ((7x+1)=0) से \(-\frac{1}{7}\) मिलता है। / ((7x+1)(x+1)=0), so \(x=-\frac{1}{7}\) and (-1). In exams, ((7x+1)=0) gives \(-\frac{1}{7}\).
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What are the roots of \(x^2+2x-2=0\) by the quadratic formula?
A. \(x=-1\pm\sqrt{3}\)
यहां (D=22-4(1)(-2)=12), इसलिए \(x=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt{3}\) है। परीक्षा में \(\sqrt{12}=2\sqrt{3}\) सरल करें। / Here (D=22-4(1)(-2)=12), so \(x=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt{3}\). In exams, simplify \(\sqrt{12}=2\sqrt{3}\).
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Which factorised form is correct for \(3x^2-10x-8=0\)?
A. ((3x+2)(x-4)=0)
((3x+2)(x-4)=3x-2-10x-8), इसलिए यह सही गुणनखंड रूप है। परीक्षा में विस्तार करके उत्तर जांचें। / ((3x+2)(x-4)=3x-2-10x-8), so this is the correct factorised form. In exams, check the answer by expanding.
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What are the roots of \(3x^2-10x-8=0\)?
A. \(x=4,-\frac{2}{3}\)
((3x+2)(x-4)=0), इसलिए \(x=-\frac{2}{3}\) और (4) हैं। परीक्षा में संकेत बदलकर मूल लिखें। / ((3x+2)(x-4)=0), so \(x=-\frac{2}{3}\) and (4). In exams, change signs while writing roots.
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If a student wrote only (x=7) from \(x^2=49\), what is the correct correction?
A. \(x=\pm7\) लिखना चाहिएOne should write \(x=\pm7\)
\(x^2=49\) से \(x=\pm\sqrt{49}=\pm7\) मिलता है। परीक्षा में वर्गमूल विधि में दोनों चिन्ह अनिवार्य हैं। / From \(x^2=49\), \(x=\pm\sqrt{49}=\pm7\). In exams, both signs are necessary in the square root method.
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If \(x^2-6x-7=0\) is solved by completing the square, which middle step is correct?
A. ((x-3)2=16)
\(x^2-6x=7\) में (9) जोड़ने पर ((x-3)2=16) मिलता है। परीक्षा में आधे गुणांक का वर्ग दोनों पक्षों में जोड़ें। / Adding (9) to \(x^2-6x=7\) gives ((x-3)2=16). In exams, add the square of half the coefficient to both sides.
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What roots are obtained for \(x^2-6x-7=0\) by completing the square method?
A. (x=7,-1)
((x-3)2=16), इसलिए \(x-3=\pm4\) और (x=7,-1) हैं। परीक्षा में \(\pm\) से दोनों मूल निकालें। / Since ((x-3)2=16), \(x-3=\pm4\), so (x=7,-1). In exams, use \(\pm\) to find both roots.
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