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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 4 · Set 2 · Q24
α, β are zeroes of the polynomial p (x)=3 x^2-6 x-5. Find the value of 1/(α^2)+1/(β^2).
$\displaystyle \alpha, \beta$ are zeroes of the polynomial $\displaystyle \mathrm{p}(x)=3 x^{2}-6 x-5$. Find the value of $\displaystyle \frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$.
Marking-scheme solution
\[\begin{aligned}
& \alpha+\beta=2, \alpha \beta=-\frac{5}{3} \\
& \therefore \frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}=\frac{4+\dfrac{10}{3}}{\dfrac{25}{9}} \\
& =\frac{66}{25}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.