Question

Find the value of $-k$:

If $\alpha$ and $\beta$ are the zeros of the polynomial $x^2-6x+k$ such that $3\alpha +2\beta =20$

Answer 1

Given,
$x^2-6x+k$
Where,
$\alpha +\beta =\frac{-b}{a}=-\frac{-6}{1}=6$
$\alpha \beta =\frac{c}{a}=\frac{k}{1}=k$
We have,
$3\alpha +2\beta =20$
$=>\alpha +2\alpha +2\beta =20$
$=>\alpha +2(\alpha +\beta )=20$
$=>\alpha +2\left(6\right)=20$
$=>\alpha =20-12$
$=>\alpha =8$
Putting the value in equation as $\alpha$ is the zero of the polynomial,we get,
$x^2-6x+k$
$=(8{)}^2-6(8)+k$
$=64-48+k$
$=16+k$
$=>k=-16$