Question

The roots of the equation, \(6x^4-25x^3+12x^2+25x+6=0\) are

Answer 1

Given; \(6x^4-25x^3+12x^2+25x+6=0\)
Dividing the equation by \(x^2,\) we get:
\(6\bigg(x^2+\dfrac{1}{x^2}\bigg)-25\bigg(x-\dfrac{1}{x}\bigg)+12=0\)
Let: \(x-\dfrac{1}{x}=t\)
\(\Rightarrow x^2-1=xt\)
\(\Rightarrow x^2-xt-1=0\)
Now, this equation should have real roots:
\(\Rightarrow D\ge0\Rightarrow t^2+4\ge0,\) which is true \(\forall ~t\in\mathbb{R} \)
Now, Squaring both sides, we get:
\(x^2+\dfrac{1}{x^2}=t^2+2\)
Substituing both we get the above equation as:
\(6t^2-25t+24=0\)
\(\Rightarrow 6t^2-9t-16t+24=0\)
\(\Rightarrow 3t(2t-3)-8(2t-3)=0\)
\(\Rightarrow (3t-8)(2t-3)=0\)
\(\Rightarrow t=\dfrac{8}{3}, \dfrac{3}{2}\)

If \(t=\dfrac{8}{3}\), we have
\(x-\dfrac{1}{x}=\dfrac{8}{3}\)
\(\Rightarrow 3(x^2-1)=8x\)
\(\Rightarrow 3x^2-8x-3=0\)
\(\Rightarrow 3x(x-3)+1(x-3)=0\)
\(\Rightarrow (x-3)(3x+1)=0\)
\(\Rightarrow x=-\dfrac{1}{3},3\)

If \(t=\dfrac{3}{2}\), we have
\(x-\dfrac{1}{x}=\dfrac{3}{2}\)
\(\Rightarrow 2(x^2-1)=3x\)
\(\Rightarrow 2x^2-2-3x=0\)
\(\Rightarrow 2x^2-4x+x-2=0\)
\(\Rightarrow (2x+1)(x-2)=0\)
\(\Rightarrow x=-\dfrac{1}{2},2\)

\(\therefore\) All possible values of \(x\) are \(-\dfrac{1}{2},2,-\dfrac{1}{3},3\)