Question

If $x=\frac{2}{3}$ and $x=-3$ are the roots of the equation $a{x}^2+7x+b=0,$ find the value of $a^2+b^2$.

Answer 1

$\Rightarrow$ $x=\frac{2}{3}$ and $x=-3$ are the roots of the equation, then $(x-\frac{2}{3})(x-(-3\left)\right)=0$ be that quadratic equation.

$\Rightarrow$ $(x-\frac{2}{3})(x-(-3\left)\right)=0$

$\Rightarrow$ $(x-\frac{2}{3})(x+3)=0$

$\Rightarrow$ $x(x+3)-\frac{2}{3}(x+3)=0$

$\Rightarrow$ $x^2+3x-\frac{2}{3}x-2=0$

$\Rightarrow$ $3x^2+9x-2x-6=0$

$\Rightarrow$ $3x^2+7x-6=0$

$\Rightarrow$ $3x^2+7x-6=0$, comparing it with $a{x}^2+7x+b=0$

$\therefore$ $a=3$ and $b=-6$

$\Rightarrow$ $a^2+b^2=(3{)}^2+(-6{)}^2=9+36=45$