Question

Which of the following value of $x$ satisfies the equation ?

$12(x^2+\frac{1}{x^2})-56(x+\frac{1}{x})+89=0$

• A. $-\frac{2}{3}$
• B. $-\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

Answer: (C), (D)

$2$
$\frac{3}{2}$

$12(x^2+\frac{1}{x^2})-56(x+\frac{1}{x})+89=0$

Let $(x+\frac{1}{x})$ = m

Squaring both sides:

$x^2+\frac{1}{x^2}+2=m^2$

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

Reframing the equation,

$12(m^2-2)-56m+89=0$

$12m^2-56m+65=0$

$m=\frac{56\pm \sqrt{16}}{24}$

$m=\frac{52}{24},\frac{60}{24}$

$m=\frac{13}{6},\frac{5}{2}$

when, $m=\frac{13}{6}$

then, $x+\frac{1}{x}=\frac{13}{6}$

$6x^2-13x+6=0$

$6x^2-9x-4x+6=0$

$(3x-2)(2x-3)=0$

$x=\frac{3}{2},\frac{2}{3}$

when, $m=\frac{5}{2}$

then, $x+\frac{1}{x}=\frac{5}{2}$

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

$2x^2-4x-x+2=0$

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

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