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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 6 · Set 2 · Q29
Find the zeroes of the polynomial q (x)=8 x^2-2 x-3. Hence, find a polynomial whose zeroes are 2 less than the zeroes of q (x).
Find the zeroes of the polynomial $\displaystyle \mathrm{q}(x)=8 x^{2}-2 x-3$. Hence, find a polynomial whose zeroes are $\displaystyle 2$ less than the zeroes of $\displaystyle \mathrm{q}(x)$.
Marking-scheme solution
\[\begin{aligned}
& p(x)=8 x^{2}-2 x-3 \\
& \text { Zeroes are }-\frac{1}{2} \text { and } \frac{3}{4} \\
& \text { New zeroes are }-\frac{5}{2} \text { and }-\frac{5}{4} \\
& \text { Sum of new zeroes }=\frac{-5}{2}+\frac{-5}{4}=\frac{-15}{4} \\
& \text { Product of new zeroes }=\left(-\frac{5}{2}\right) \times\left(-\frac{5}{4}\right)=\frac{25}{8} \\
& \text { Required polynomial is } x^{2}+\frac{15}{4} x+\frac{25}{8} \text { or } 8 x^{2}+30 x+25
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.