Question

Solve the equation :
$6x^4-25x^3+12x^2+25x+6=0$

Answer 1

Solution:-

$6x^4-25x^3+12x^2+25x+6=0$

$6x^4+12x^2+6-25x^3+25x=0$

$\Rightarrow 6(x^4+2x^2+1)-25x(x^2-1)=0$

$\Rightarrow 6{(x^2+1)}^2-25x(x^2-1)=0$

$\Rightarrow 6({(x^2-1)}^2+4x^2)-25x(x^2-1)=0$

Let $x^2-1=t$

Therefore,

$6(t^2+4x^2))-25tx=0$

$6t^2+24x^2-25tx=0$

$\Rightarrow 24x^2-25tx+6t^2=0$

$\Rightarrow 24x^2-9tx-16tx+6t^2=0$

$\Rightarrow 3x(8x-3t)-2t(8x-3t)=0$

$\Rightarrow (8x-3t)(3x-2t)=0$

$\Rightarrow 8x-3t=0\text{or}3x-2t=0$

Substituting the value of $t$, we have

$\Rightarrow 8x-3(x^2-1)=0\text{or}3x-2(x^2-1)=0$

$\Rightarrow 8x-3x^2+3=0\text{or}3x-2x^2+2=0$

$\Rightarrow 3x^2-8x-3=0\text{or}2x^2-3x-2=0$

Now,

• $3x^2-8x-3=0$

$3x^2-9x+x-3=0$

$(3x+1)(x-3)=0$

$x=\frac{-1}{3}\text{or}x=3$

• $2x^2-3x-2=0$

$2x^2-4x+x-2=0$

$(x-2)(2x+1)=0$

$x=2\text{or}x=\frac{-1}{2}$

Hence for the equation $6x^4-25x^3+12x^2+25x+6=0$, the values of $x$ are $\frac{-1}{3},\frac{-1}{2},2$ and $3$.