CBSE Class 10 Mathematics Basic 2026: question 32(b)

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Step-by-step solution

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1. For $2x^2+kx+3=0$: $a=2$, $b=k$, $c=3$. For real and equal roots, $D=b^2-4ac=0$, so $k^2-4(2)(3)=0\Rightarrow k^2-24=0$.

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2. $k=\pm \sqrt{24}=\pm 2\sqrt{6}$

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3. The equations obtained are $2x^2+2\sqrt{6}x+3=0$ and $2x^2-2\sqrt{6}x+3=0$.

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4. For $k=2\sqrt{6}$: $(\sqrt{2}x+\sqrt{3}{)}^2=0$, so the equal roots are $x=-\frac{b}{2a}=-\frac{2\sqrt{6}}{4}=-\frac{\sqrt{6}}{2}=-\sqrt{\frac{3}{2}}$, i.e. $-\sqrt{\frac{3}{2}},-\sqrt{\frac{3}{2}}$.

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5. For $k=-2\sqrt{6}$: $(\sqrt{2}x-\sqrt{3}{)}^2=0$, so the equal roots are $x=\frac{2\sqrt{6}}{4}=\sqrt{\frac{3}{2}}$, i.e. $\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}$.

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Answer: k = ±2√6; roots −√(3/2), −√(3/2) (for k = 2√6) and √(3/2), √(3/2) (for k = −2√6)

Revise the chapter

• Chapter: Quadratic Equations (notes, formulas, where marks are lost and every past question)
• Topic: Nature of Roots and Discriminant