Question

Find the roots of the following equation:

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

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

Answer 1

Answer: (B)

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

$2(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})+14=0$
Let $x+\frac{1}{x}=y$.........(1)
then, $x^2+\frac{1}{x^2}=y^2$
$x^2+\frac{1}{x^2}=y^2-2$
hence, $2(y^2-2)-9y+14=0$
$2y^2-4-9y+14=0$
$2y^2-9y+10=0$
$2y^2-5y-4y+10=0$
$(y-2)(2y-5)=0$
$y=\frac{5}{2},2$
Now substituting value of $y$ in equation (1), we get

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