Question

Find the value of
$(x^2+\frac{1}{x^2})-4(x+\frac{1}{x})+6=0$

Answer 1

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

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

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

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

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

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

$\Rightarrow x+\frac{1}{x}-2=0$

$\Rightarrow x^2-2x+1=0$ is of the form $a^2-2ab+b^2={(a-b)}^2$

${(x-1)}^2=0$

$\therefore x=1$