If $\alpha, \beta$ be the zeroes of the polynomial $2 \mathrm{x}^{2}+5 \mathrm{x}+\mathrm{k}$ such that $(\alpha+\beta)^{2}-\alpha \beta=\frac{21}{4}$, then $\mathrm{k}=$ ? (a) 3 (b) $-3$ (c) -2 (d) 2

The correct option is option (d) 2

$\alpha$ and $\beta$ are the zeroes of $2 \mathrm{x}^{2}+5 \mathrm{x}+\mathrm{k}$, we have:

$\alpha+\beta=\frac{-5}{2}$ and $\alpha \beta=\frac{k}{2}$

it is given that $\alpha^{2}+\beta^{2}+\alpha \beta=\frac{21}{4}$.

$\Rightarrow(\alpha+\beta)^{2}-\alpha \beta=\frac{21}{4}$

$$
\begin{aligned}
&\Rightarrow\left(\frac{-5}{2}\right)^{2}-\frac{k}{2}=\frac{21}{4} \\
&\Rightarrow \frac{25}{4}-\frac{k}{2}=\frac{21}{4} \\
&\Rightarrow \frac{k}{2}=\frac{25}{4}-\frac{21}{4}=\frac{4}{4}=1 \\
&\Rightarrow \mathrm{k}=2
\end{aligned}
$$