Question

$6(x^2+\frac{1}{x^2})-25(x-\frac{1}{x})+12=0$

• A. , , 2, 3
• B. , , -2, -3
• C. , , 2, 3
• D. , , -2, 3
• E. , 2, -3

Answer 1

The correct option is A , , 2, 3

Soln:
Let$=z$
$(\frac{x-1}{x}{)}^2=z^2$
$(x^2+\frac{1}{x^2}-2)=z^2$
$x^2+\frac{1}{x^2}=z^2+2$
Now substituting in the equation:
$6(z^2-2)-25z+12=0$
$6z^2+12-25z+12=0$
$6z^2-25z+24=0$
$6z^2-16z-9z+24=0$
$2z(3z-8)-3(3a-8)=0$
$(2z-3)(3z-8)=0$
$z=\frac{3}{2};z=\frac{8}{3}$
$x-\frac{1}{x}=\frac{3}{2}$
$x-\frac{1}{x}=\frac{8}{3}$
$2x^2=3x-2=0$
$3x^2-8x-3=0$
$2x^2-4x+x-2=0$
$3x^2-9x+x-3=0$
$x(x-2)+1(x-2)=0$
$3x(x-3)+1(x-3)=0$
$(2x+1)(x-2)=0$
$(3x+1)(x-3)=0$
$x=-\frac{1}{2},x=2$
$x=-\frac{1}{3},x=3$
So,the roots are:$-\frac{1}{2},2,-\frac{1}{3},3$
Hence,Option (a).