# If α, β be the zero of the polynomial

Question:

If $\alpha, \beta$ be the zero of the polynomial $2 x^{2}+5 x+k$ such that $\alpha^{2}+\beta^{2}+\alpha \beta=\frac{21}{4}$, then $k=?$

(a) 3
(b) −3
(c) −2
(d) 2

Solution:

$(\mathrm{d}) 2$

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

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

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

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

$=>\left(\frac{-5}{2}\right)^{2}-\frac{k}{2}=\frac{21}{4}$

$=>\frac{25}{4}-\frac{k}{2}=\frac{21}{4}$

$=>\frac{k}{2}=\frac{25}{4}-\frac{21}{4}=\frac{4}{4}=1$

$=>k=2$