Question

If a, b are the zeros of the polynomial $f\left(x\right)=2x^2+5x+k$ satisfying the relation $a^2+b^2+ab=\frac{21}{4}$, then find the value of k for this to be possible.

Answer 1

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

Therefore,

$a+b=-\frac{5}{2}$ and $ab=\frac{k}{2}$

Now,

$a^2+b^2+ab=\frac{21}{4}$

$(a^2+b^2+2ab)-ab=\frac{21}{4}$

$(a+b{)}^2-ab=\frac{21}{4}$

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

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

$k=2$