Question

Solve the equations:
$2x^3-x^2-22x-24=0$, two of the roots being in the ratio of $3:4$.

Answer 1

${2x}^3{-x}^2-22x-24=0$

Let the roots of the equation be $3a,4a,b$

$\sum x=3a+4a+b=-\frac{-1}{2}=\frac{1}{2}\Rightarrow b=\frac{1}{2}-7a....\left(i\right)\sum xy=3a\left(4a\right)+4a\left(b\right)+b\left(3a\right)=\frac{-22}{2}7a^2+7ab=-11$

Substituitng $b$ from $\left(i\right)$

$7a^2+7a(\frac{1}{2}-7a)=-1174a^2-7a-22=074a^2+37a-44a-22=0\Rightarrow a=-\frac{1}{2},\frac{22}{37}$

Substituting $a$ in $\left(i\right)$

$\Rightarrow b=4,\frac{-40}{37}$

$\sum xyz=3a\left(4a\right)b=-\frac{-24}{2}\Rightarrow a^{2}b=1$

The product of roots do not satisfy $a=\frac{22}{37}$

So the roots of the equation are $\frac{-3}{2},-2,4$