Question

If two roots of the equation $4x^2-2x-3=0$ are a and b, then find out the value of $\frac{a^2}{b}+\frac{b^2}{a}.$

Answer 1

$4x^2-2x-3=0$
Let a and b be the roots (we cannot type alpha and beta here, hence we call it a and b)
$a+b=\frac{2}{4}=\frac{1}{2}$
$ab=\frac{-3}{4}$
$\frac{a^2}{b}+\frac{b^2}{a}=\frac{(a^3+b^3)}{ab}$ ...(I)
Now you know the identiry $a^3+b^3=(a+b{)}^3-3ab(a+b)$
Substituting on RHS:
$a^3+b^3=(1{)}^3-3\left(\frac{-3}{4}\right)\left(\frac{1}{2}\right)$
$=\frac{1}{8}+\frac{9}{8}=\frac{10}{8}$
Substituting in (i),
$\frac{a^2}{b}+\frac{b^2}{a}=\frac{\left(\frac{10}{8}\right)}{\left(\frac{-3}{4}\right)}$
$=\frac{-40}{24}=\frac{-5}{3}=-1.6667$