Question

If $\frac{2}{3}$ and$-3$ are the roots of equation $p{x}^2+7x+q=0$, find $p,q$

Answer 1

If $x=\frac{2}{3},x=-3$ are roots of the equation $p{x}^2+7x+q=0$ then

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

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

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

$\Rightarrow x^2+\frac{7}{3}x-2=0$

$\Rightarrow 3x²+7x-6=0$

Compare the equation with $px²+7x+q=0$

Therefore, $p=3,q=-6$