Question

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

• A. $\frac{3}{2},-2,-\frac{2}{3},3$
• B. $\frac{1}{2},-2,\frac{1}{3},-3$
• C. $-\frac{1}{2},2,-\frac{1}{3},3$
• D. $\frac{3}{2},2,\frac{2}{3},3$

Answer 1

The correct option is C $-\frac{1}{2},2,-\frac{1}{3},3$

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

If $t=\frac{8}{3}$, we have
$x-\frac{1}{x}=\frac{8}{3}$

$\Rightarrow 3x(x-3)+1(x-3)=0$
$\Rightarrow (x-3)(3x+1)=0$
$\Rightarrow x=-\frac{1}{3},3$

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

$\therefore$ All possible values of $x$ are $-\frac{1}{2},2,-\frac{1}{3},3$