If $\alpha$ and $\beta$ are the zeros of the polynomial $p(x)=2 x^{2}+5 x+k$ satisfying the relation $\alpha^{2}+\beta^{2}+\alpha \beta=\frac{21}{4}$ then find the value of $k$.
Let $\alpha$ and $\beta$ be the zeroes of the polynomial $p(x)=2 x^{2}+5 x+k$.
Sum of zeroes $=-\frac{b}{a}$
$\Rightarrow \alpha+\beta=-\frac{5}{2} \quad \ldots(1)$
and
Product of zeroes $=\frac{c}{a}$
$\Rightarrow \alpha \beta=\frac{k}{2} \quad \ldots(2)$
Now, using (1)
$\alpha+\beta=-\frac{5}{2}$
Squaring both sides, we get
$\Rightarrow(\alpha+\beta)^{2}=\left(-\frac{5}{2}\right)^{2}$
$\Rightarrow \alpha^{2}+\beta^{2}+2 \alpha \beta=\frac{25}{4}$
$\Rightarrow \alpha^{2}+\beta^{2}+\alpha \beta+\alpha \beta=\frac{25}{4}$
$\Rightarrow \frac{21}{4}+\alpha \beta=\frac{25}{4}$
$\Rightarrow \alpha \beta=\frac{25}{4}-\frac{21}{4}$
$\Rightarrow \alpha \beta=1$
$\Rightarrow \frac{k}{2}=1 \quad($ from $(2))$
$\Rightarrow k=2$
Hence, the value of k is 2.