Question

Show that $x^5-5x^3+5x^2-1=0$ has three equal roots and find that root.

Answer 1

Given: $x^5-5x^3+5x^2-1$

$=x^5-1x^3-4x^3+4x^2+x^2-1$

$=x^3(x^2-1)-4x^2(x-1)+1(x^2-1)$

$=(x-1)\left[x^3\right(x+1)-4x^2+1(x+1\left)\right]$ [Using identity, $a^2-b^2=(a-b)(a+b)$]

$=(x-1)(x^4+x^3-4x^2+x+1)$

$=(x-1)(x^4-x^3+2x^3-2x^2-2x^2+2x-x+1)$

$=(x-1)\left[x^3(x-1)+2x^2(x-1)-2x(x-1)-1(x-1)\right]$

$=(x-1{)}^2[x^3+2x^2-2x-1]$

$=(x-1{)}^2[x^3-x^2+3x^2-3x+x-1]$

$=(x-1{)}^2[x^2(x-1)+3x(x-1)+1(x-1)]$

$=(x-1{)}^3(x^2+3x+1)$

Since, $x^5-5x^3+5x^2-1=0$,

$\Rightarrow (x-1{)}^3(x^2+3x+1)=0$

$\Rightarrow (x-1{)}^3=0$ or $x^2+3x+1=0$

$\Rightarrow x=1$ or $\frac{-3\pm \sqrt{5}}{2}$

Thus, the given equation has three equal roots which is $1$ and two distinct roots which is $\frac{-3\pm \sqrt{5}}{2}$.