Question

Solve for $x:2(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})+14=0$

• A. $\frac{1}{2},1,2$
• B. $\frac{1}{2},1,-2$
• C. $\frac{1}{2},1,4$
• D. $\frac{1}{2},1,-4$

Answer 1

The correct option is A $\frac{1}{2},1,2$

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

$\therefore x^2+\frac{1}{x^2}+2=t^2$

$\therefore x^2+\frac{1}{x^2}=t^2-2$

Substituting these in the equation, we get

$2(t^2-2)-9t+14=0$

$\therefore 2t^2-9t+10=0$

$\therefore (t-2)(2t-5)=0$

$\therefore t=2$ or $t=\frac{5}{2}$

Case 1) $t=2$

$\therefore x+\frac{1}{x}=2$

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

$\therefore x=1$

Case 2) $t=\frac{5}{2}$

$\therefore x+\frac{1}{x}=\frac{5}{2}$

$\therefore 2x^2-5x+2=0$

$\therefore (x-2)(2x-1)=0$

$\therefore x=2$ or $x=\frac{1}{2}$

$\therefore x=\{\frac{1}{2},1,2\}$