Question

If $\alpha$, $\beta$ are the zeros of the polynomial $f\left(x\right)=2x^2+5x+k$ satisfying the relation ${\alpha }^2+{\beta }^2+\alpha \beta =\frac{21}{4}$, then find the value of k for this condition to be true. [3 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 1 Mark

Since $\alpha$ and $\beta$ are the zeros of the polynomial $f\left(x\right)=2x^2+5x+k$

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

Now,

${\alpha }^2+{\beta }^2+\alpha \beta =\frac{21}{4}$

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

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

$\Rightarrow \frac{25}{4}-\frac{k}{2}=\frac{21}{4}$ $[\because \alpha +\beta =-\frac{5}{2}and\alpha \beta =\frac{k}{2}]$

$\Rightarrow -\frac{k}{2}=-1$

$\Rightarrow k=2$