CBSE 2025 · Region 3 · Set 2 · Q21 · 2 marks
If the sum of the zeroes of the polynomial $\displaystyle \mathrm{p}(\mathrm{x})=(\mathrm{p}+1) \mathrm{x}^{2}+(2 \mathrm{p}+3) \mathrm{x}+(3 \mathrm{p}+4)$ is -$\displaystyle 1$, then find the value of 'p'.If $\displaystyle \alpha$ and $\displaystyle \beta$ are zeroes of the polynomial $\displaystyle \mathrm{p}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}-1$, then find the value of $\displaystyle \frac{1}{2 \alpha}+\frac{1}{2 \beta}+3 \alpha \beta$.
If the sum of the zeroes of the polynomial $\displaystyle \mathrm{p}(\mathrm{x})=(\mathrm{p}+1) \mathrm{x}^{2}+(2 \mathrm{p}+3) \mathrm{x}+(3 \mathrm{p}+4)$ is -$\displaystyle 1$, then find the value of 'p'.
If $\displaystyle \alpha$ and $\displaystyle \beta$ are zeroes of the polynomial $\displaystyle \mathrm{p}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}-1$, then find the value of $\displaystyle \frac{1}{2 \alpha}+\frac{1}{2 \beta}+3 \alpha \beta$.
Marking-scheme solution
Sum of zeroes \(\displaystyle =-\frac{2 \mathrm{p}+3}{\mathrm{p}+1}=-1\)
\(\displaystyle \mathrm{p}=-2\)
\[\begin{aligned}
& \alpha+\beta=2 \\
& \alpha \beta=-1 \\
& \frac{1}{2 \alpha}+\frac{1}{2 \beta}+3 \alpha \beta=\frac{\alpha+\beta}{2 \alpha \beta}+3 \alpha \beta \\
& =\frac{2}{2(-1)}+3(-1)=-4
\end{aligned}
\]
PolynomialsRelationship between Zeroes and CoefficientsApplyvery_short_answermedium
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.