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

Answer: (A)

$\frac{1}{2},1,2$

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

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

So, the above equation would become

$2(k^2-2)-9k+14=02k^2-9k+10=02k^2-4k-5k+10=02k(k-2)-5(k-2)=0(2k-5)(k-2)=0k=2,\frac{5}{2}$

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

$x+\frac{1}{x}=\frac{5}{2}{x}^2-\frac{5x}{2}+1=02x^2-5x+2=02x^2-4x-x+2=02x(x-2)-1(x-2)=0(2x-1)(x-2)=0x=\frac{1}{2},2$

So, $x$ equals $1,\frac{1}{2},2$.