Question

Solve the equations:
$24x^3+46x^2+9x-9=0$, one root being double another of the roots.

Answer 1

${24x}^3{+46x}^2+9x-9=0$

Let the roots be $a.2a,b$

$\sum x=a+2a+b=-\frac{46}{24}\Rightarrow b=-\frac{23}{12}-3a......\left(i\right)\sum xy=a\left(2a\right)+2a\left(b\right)+b\left(a\right)=\frac{9}{24}=\frac{3}{8}\Rightarrow 2a^2+3ab=\frac{3}{8}$

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

$\Rightarrow 2a^2+3a(-\frac{23}{12}-3a)=\frac{3}{8}\Rightarrow 56a^2+46a+3=0\Rightarrow 56a^2+4a+42a+3=0\Rightarrow a=-\frac{3}{4},\frac{-1}{14}$

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

$\Rightarrow b=\frac{1}{3},-\frac{143}{84}\sum xyz=2a^{2}b=-\frac{-9}{24}=\frac{3}{8}\Rightarrow 2a^{2}b=\frac{3}{8}$

Product of roots is not satisfied for $a=-\frac{1}{14}$

So the roots of the equation are $-\frac{3}{4},-\frac{3}{2},\frac{1}{3}$