Question

Solve the equations:
$4x^3+20x^2-23x+6=0$, two of the roots being equal.

Answer 1

${4x}^3+{20x}^2-23x+6=0$

Let the roots be $a,a,b$

$\sum x=a+a+b=-\frac{20}{4}=-52a+b=-5\Rightarrow b=-5-2a....\left(i\right)\sum xy=a\left(a\right)+a\left(b\right)+b\left(a\right)=\frac{-23}{4}{a}^2+2ab=\frac{-23}{4}$

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

$4a^2+8ab+23=04a^2+8a(-5-2a)+23=04a^2-40a-16a^2+23=012a^2+40a-23=012a^2-6a+46a-23=0\Rightarrow a=\frac{1}{2},-\frac{23}{6}$

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

$\Rightarrow b=-6,-\frac{8}{3}$

$\sum xyz=a^{2}b=-\frac{6}{4}=-\frac{3}{2}$

Product of roots do not satisfy $a=-\frac{8}{3}$

So the roots of the equation are $\frac{1}{2},\frac{1}{2},-6$