Question

If $2(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})+14=0$ then x = ____________

• A. $\frac{1}{2}$
• B. 1
• C. 2
• D. All of these

Answer 1

The correct option is D All of these

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

$\Rightarrow$ $2[{(x+\frac{1}{x})}^2-2\times x\times \frac{1}{x}]-9(x+\frac{1}{x})+14=0$

$\Rightarrow$ $2{(x+\frac{1}{x})}^2-4-9(x+\frac{1}{x})+14=0$

$\Rightarrow$ $2{(x+\frac{1}{x})}^2-9(x+\frac{1}{x})+10=0$

$\Rightarrow$ $2{(x+\frac{1}{x})}^2-5(x+\frac{1}{x})-4(x+\frac{1}{x})+10=0$

$\Rightarrow$ $[2(x+\frac{1}{x})-5][(x+\frac{1}{x})-2]=0$

$2(x+\frac{1}{x})-5=0$ $or$ $x+\frac{1}{x}-2=0$

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

$\Rightarrow$ $\frac{x^2+1}{x}=\frac{5}{2}$ $\Rightarrow$ $\frac{x^2+1}{x}=2$

$\Rightarrow$ $2x^2-5x+2=0$ $\Rightarrow$ $x^2-2x+1=0$

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

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

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

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

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