Question

Solve the following equation that has equal roots:
${x^6} - 2{x^5} - 4{x^4} + 12{x^3} - 3{x^2} - 18x + 18 = 0$

Answer 1

Hint: Here we will simplify the given equation into simpler form and by using the determinant formula the roots can be calculated.

Complete step-by-step answer:

Given equation is ${x^6} - 2{x^5} - 4{x^4} + 12{x^3} - 3{x^2} - 18x + 18 = 0$
Let $f(x) = {x^6} - 2{x^5} - 4{x^4} + 12{x^3} - 3{x^2} - 18x + 18$
Differentiate above equation w.r.t x
$\Rightarrow \dfrac{d}{{dx}}f(x) = f'(x) = 6{x^5} - 10{x^4} - 16{x^3} + 36{x^2} - 6x - 18$
Now factorize f(x)
\Rightarrow f(x) = {({x^2} - 3)^2}({x^2} - 2x + 2) = 0 \\\ {({x^2} - 3)^2} = 0{\text{ & }}({x^2} - 2x + 2) = 0 \\\ \Rightarrow x = \pm \sqrt 3 \\\
Second equation is quadratic equation so apply quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow \dfrac{{2 \pm \sqrt {4 - 8} }}{2} = \dfrac{{2 \pm \sqrt {4{i^2}} }}{2}[\because {i^2} = - 1] \\\ \Rightarrow \dfrac{{2 \pm 2i}}{2} = 1 \pm i \\\$
Put $x = \sqrt 3$ in $f'(x)$
$\Rightarrow f'(x) = 6{(\sqrt 3 )^5} - 10{(\sqrt 3 )^4} - 16{(\sqrt 3 )^3} + 36{(\sqrt 3 )^2} - 6\sqrt 3 - 18 \\\ \Rightarrow f'(x) = 54\sqrt 3 - 90 - 48\sqrt 3 + 108 - 6\sqrt 3 - 18 \\\ \Rightarrow f'(x) = 0 \\\$
Put $x = - \sqrt 3$ in $f'(x)$
$\Rightarrow f'(x) = 6{( - \sqrt 3 )^5} - 10{( - \sqrt 3 )^4} - 16{( - \sqrt 3 )^3} + 36{( - \sqrt 3 )^2} + 6\sqrt 3 - 18 \\\ \Rightarrow f'(x) = - 54\sqrt 3 - 90 + 48\sqrt 3 + 108 + 6\sqrt 3 - 18 \\\ \Rightarrow f'(x) = 0 \\\$
Therefore, ${({x^2} - 3)^2}$ is the HCF of f(x) and $f'(x)$.
Hence $x = \pm \sqrt 3$ is a double root of f(x) =0
Given equation has equal roots and the roots of the equation have $x = \pm \sqrt 3 ,1 \pm i$.
So this is your desired answer.

Note: In this type of question if the equation has equal roots then its differentiation is zero at that root i.e the equation has a double root.