Question

Solve the equations:
$8x^4-2x^3-27x^2+6x+9=0$, two of the roots being equal but opposite in sign.

Answer 1

$8x^4-2x^3-27x^2+6x+9=0$

Let the roots be $a,-a,b,c$

$S_1=a-a+b+c=-\frac{-2}{8}=\frac{1}{4}b+c=\frac{1}{4}.....\left(i\right)$

$S_3=-a^{2}b-abc+abc-a^{2}c=-\frac{6}{8}\Rightarrow a^2(b+c)=\frac{3}{4}.....\left(ii\right)$

dividing $\left(i\right)$ by $\left(ii\right)$

$\Rightarrow b+c=\frac{1}{4}......\left(iii\right)S_4=-a^{2}bc=\frac{9}{8}\Rightarrow bc=-\frac{3}{8}....\left(iv\right)$

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

$b(\frac{1}{4}-b)=-\frac{3}{8}\frac{b-4b^2}{4}=-\frac{3}{8}2b-8b^2=38b^2-2b-3=08b^2+4b-6b-3=04b(2b+1)-3(2b+1)=0(4b-3)(2b+1)=0\Rightarrow b=\frac{3}{4},-\frac{1}{2}$

substituing $b$ in $\left(iv\right)$

$\Rightarrow c=-\frac{1}{2},\frac{3}{4}$

$S_2=a(-a)+(-a)\left(b\right)+bc+ca=\frac{-27}{8}\Rightarrow -a^2-ab+bc+ac=\frac{-27}{8}$

The value of $s_4$ is not satisfied for $b=-\frac{1}{2}$

So the roots of the equation are

$\sqrt{3},-\sqrt{3},\frac{3}{4},-\frac{1}{2}$