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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 1 · Set 2 · Q21
If the zeroes of the polynomial x^2+ ax + b are in the ratio 3: 4, then prove that 12 a^2=49 b.
If the zeroes of the polynomial $\displaystyle \mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}$ are in the ratio $\displaystyle 3$ : $\displaystyle 4$, then prove that $\displaystyle 12 \mathrm{a}^{2}=49 \mathrm{~b}$.
Marking-scheme solution
Let the zeroes are \(\displaystyle 3 \alpha\) and \(\displaystyle 4 \alpha\)
\[\begin{aligned}
& 3 \alpha+4 \alpha=-\mathrm{a} \\
& \Rightarrow 7 \alpha=-\mathrm{a} \\
& \text { Also, } 12 \alpha^{2}=\mathrm{b} \\
& \text { LHS }=12 \mathrm{a}^{2}=12(-7 \alpha)^{2}=49 \times 12(\alpha)^{2}=49 \mathrm{~b}=\text { RHS }
\end{aligned}
\]
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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.