Question

Solve the following equations which have equal roots:
$x^5-x^3+4x^2-3x+2=0$.

Answer 1

Give equation, $x^5-x^3+4x^2-3x+2=0$

Consider $f\left(x\right)=x^5-x^3+4x^2-3x+2$

For $x=-2,f\left(x\right)=0⟹x=-2$ is one root of the given equation.

$f\left(x\right)=(x+2)(x^4-2x^3+3x^2-2x+1)$

Second factor in above has imaginary roots. We know that $f\left(x\right)=0$ has equal roots and since imaginary roots occurs as conjugates, we can assume that the five roots of $f\left(x\right)=0$ are $(a+ib),(a+ib),(a-ib),(a-ib)$

Sum of the roots $=4a=2⟹a=\frac{1}{2}$

Also, product of the roots $=(a^2+b^2{)}^2=1⟹b=\frac{\sqrt{3}}{2}$

$\therefore$ Roots of the given equation are $-2,\frac{1-\sqrt{3}i}{2},\frac{1-\sqrt{3}i}{2},\frac{1+\sqrt{3}i}{2},\frac{1+\sqrt{3}i}{2}$