Question

Solve $12x^4-56x^3+89x^2-56x+12=0?$

Answer 1

Given

$12x^4-56x^3+89x^2-56x+12=0$ ----- (1)

on dividing equation (1) by $x^2$

$\Rightarrow 12x^2-56x+89-\frac{56}{x}+\frac{12}{x^2}=0$

$12[x^2+\frac{1}{x^2}]-56[x+\frac{1}{x}]+89=0$

$\Rightarrow 12\left[\right(x+\frac{1}{x}{)}^2-2]-56[x+\frac{1}{x}]+89=0$

$\Rightarrow 12[x+\frac{1}{x}{]}^2-56[x+\frac{1}{x}]+89-24=0$

$\Rightarrow 12[x+\frac{1}{x}{]}^2-56[x+\frac{1}{x}]+65=0$

Let $x+\frac{1}{x}=y$

$12y^2-56y+65=0$

$y=\frac{56+\sqrt{(56{)}^2-48\times 65}}{24}=\frac{13}{6},\frac{5}{2}$

Now $x+\frac{1}{x}=\frac{13}{6}$ or $x+\frac{1}{x}=\frac{5}{2}$

$6x^2-13x+6=0$ or $2x^2-5x+2=0$

$x=\frac{13\pm \sqrt{169-144}}{12}$ or $x=\frac{5\pm \sqrt{25-16}}{4}$

$x=\frac{13+5}{12}$ or $x=\frac{5+3}{4}$

Hence $[x=\frac{3}{2}]$, or $\left[\frac{2}{3}\right]$ $[x=2,\frac{1}{2}]$