Question

Solve:
$30(x^2+\frac{1}{x^2})-77(x-\frac{1}{x})-12=0$

Answer 1

$30(x^2+\frac{1}{x^2})-77(x-\frac{1}{x})-12=0$
$\because$ $x^2+\frac{1}{x^2}=[{(x-\frac{1}{x})}^2+2]$
$\therefore$ $30x^2+\frac{1}{x^2}=[{(x-\frac{1}{x})}^2+2]-7777(x-\frac{1}{x})-12=0$
Let $x-\frac{1}{x}=m$
$30[m^2+2]-77-12=0$
$30m^2-77m+48=0$
$30m^2-45m-32m+48=0$
$15m(2m-3)-16(2m-3)=0$
$(2m-3)(15m-16)=0$
$2m-3=0$or $15m-16=0$
$2m=3$ or $15m=16$
$\therefore$ $m=\frac{3}{2}$ or $m=\frac{16}{15}$
By resubstituting
$m=x-\frac{1}{x}$
$\therefore$ $x-\frac{1}{x}=\frac{3}{2}$ or $x-\frac{1}{x}=\frac{16}{15}$
$x^2-1=\frac{3x}{2}$ or $x^2-1=\frac{16x}{15}$
$2x^2-2=3x$ or $15x^2-15=16x$
$2x^2-3x-2=0$ or $15x^2-16x-15=0$
$2x^2-4x+x-2=0$ or $15x^2-25x+9x-15=0$
$2x(x-2)+1(x-2)=0$ or $5(3x-5)+3(3x-5)=0$
$(x-2)(2x+1)=0$ or $(3x-5)(5x+3)=0$
$(x-2)=0$ or $3x-5=0$
$x=2$ or $3x=5$
and $2x+1=0$ or $x=\frac{5}{3}$
$2x=-1$ or $x=\frac{5}{3}$
$2x=-1$ and
$x=-\frac{1}{2}$ or $5x+3=0$
$5x=-3$
$x=-3/5$
$\therefore$ The roots of the given equation are $2,-\frac{1}{2},\frac{5}{3},-\frac{3}{5}$