If α and β are the zeros of the polynomial

Solution:

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.