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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 4 · Set 1 · Q29
Find the zeroes of the polynomial $\displaystyle 4 \mathrm{x}^{2}+4 \mathrm{x}-3$ and verify the relationship between zeroes and coefficients of the polynomial.If $\displaystyle \alpha$ and $\displaystyle \beta$ are the zeroes of the polynomial $\displaystyle \mathrm{x}^{2}+\mathrm{x}-2$, then find the value of $\displaystyle \frac{\alpha}{\beta}+\frac{\beta}{\alpha}$.
Find the zeroes of the polynomial $\displaystyle 4 \mathrm{x}^{2}+4 \mathrm{x}-3$ and verify the relationship between zeroes and coefficients of the polynomial.
If $\displaystyle \alpha$ and $\displaystyle \beta$ are the zeroes of the polynomial $\displaystyle \mathrm{x}^{2}+\mathrm{x}-2$, then find the value of $\displaystyle \frac{\alpha}{\beta}+\frac{\beta}{\alpha}$.
Marking-scheme solution
\[\begin{aligned}
& \mathrm{P}(\mathrm{x})=4 x^{2}+4 x-3 \\
& \quad=(2 \mathrm{x}+3)(2 \mathrm{x}-1) \\
& \therefore \text { Zeroes of the polynomial are } \frac{-3}{2}, \frac{1}{2} \\
& \text { Sum of Zeroes }=\frac{-3}{2}+\frac{1}{2}=\frac{-3+1}{2}=-1=\frac{-4}{4}=\frac{-(\text { coefficient of } \mathrm{x})}{\left(\text { coefficient of } \mathrm{x}^{2}\right)} \\
& \text { Product of Zeroes }=\frac{-3}{2} \times \frac{1}{2}=\frac{-3}{4}=\frac{\text { constant term }}{\text { coefficient of } \mathrm{x}^{2}}
\end{aligned}
\]
Here \(\displaystyle \alpha+\beta=-1\) and \(\displaystyle \alpha \beta=-2\)
\[\begin{aligned}
\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \beta} & =\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta} \\
& =\frac{(-1)^{2}-2(-2)}{-2}=-\frac{5}{2}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.