Question

From the quadratic equation whose roots are $\frac{3}{4}$ and $\frac{-2}{3}$.

Answer 1

Roots area $\frac{3}{4}$ and $\frac{-2}{3}$

$\therefore$ equation :

$(x-\frac{3}{4})(x+\frac{2}{3})$

$=x^2-\frac{3}{4}x+\frac{2}{3}x-\frac{2}{4}$

$=x^2-\frac{9x+8x}{12}-\frac{1}{2}$

$=x^2-\frac{x}{12}-\frac{1}{2}$

$\therefore 12x^2-x-6$ is the equation with roots 3/4 & -2/3