Question

Form a quadratic equation whose roots are $\frac{3}{4}$ and $\frac{-2}{3}$

Answer 1

Given : Roots are $\frac{3}{4}$ and $\frac{-2}{3}$

By factor theorem $(x-a)(x-b)$ is the factor of $x^2-(a+b)x+ab=0$

where $a=\frac{3}{4},b=\frac{-2}{3}$

$\therefore$ Required equation is

$x^2-(\frac{3}{4}-\frac{2}{3})x+\frac{3}{4}\times \frac{-2}{3}=0$

(or) $x^2-\left(\frac{9-8}{12}\right)x-\frac{1}{2}=0$

(or) $12x^2-x-6=0$