Question

If $2$ roots of an equation $x^3-9x^2+14x+24=0$ are in the ratio $3:2$. Find the roots.

Answer 1

$x^3-9x^2+14x+24=0$

$\because x=-1$ is $a$ factor,

$x^2-10x+24x+1)\overline{x^3-{9x}^2+14x+24}-x^3+x^2\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_{-10x}^2+14x+24{-10x}^2-10x\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_24x+2424x+24\_\_\_\_\_\_\_\_\_\_0$

$-9x^2+14x+24=(x+1)(x^2-10x+24)$

given that the roots are in ratio $3:2$

let $3K$ & $2K$ be two roots, then.

$(x-3k)(x-2k)=x^2-51x+6k$

comparing with $x^2-10x+24$

$-5k=10\Rightarrow k=2$

or $6k^2=24\Rightarrow k^2=4=k=\pm 2$

but, we have $k=2,$

$\therefore$ Roots are $6,4$

hence, roots are $1,6,4.$