Question:
If $\alpha$ and $\beta$ are the zero of $2 x^{2}+5 x-8$, then the value of $(\alpha \beta)$ is
(a) $\frac{-5}{2}$
(b) $\frac{5}{2}$
(c) $\frac{-9}{2}$
(d) $\frac{9}{2}$
Solution:
(c) $\frac{-9}{2}$
Given: $\alpha$ and $\beta$ are the zeroes of $2 x^{2}+5 x-9$
If $\alpha$ and $\beta$ are the zeroes, then $x^{2}-(\alpha+\beta) x+\alpha \beta$ is the required polynomial.
The polynomial will be $x^{2}-\frac{5}{2} x-\frac{9}{2}$.
$\therefore \alpha \beta=\frac{-9}{2}$
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