Question

If $30(x^2+\frac{1}{x^2})-77(x-\frac{1}{x})-12=0$, then $x=1,\frac{-1}{2},\frac{2}{3},\frac{-7}{5}$.

• A. Yes
• B. No
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is B No

$30(x^2+\frac{1}{x^2})-77(x-\frac{1}{x})-12=0$

Let $(x-\frac{1}{x})=m$ then $(x-\frac{1}{x}{)}^2=m^2$

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

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

Hence, reframing the equation,
$30(m^2+2)-77m-12=0$
$30m^2-77m+48=0$

$m$ = $\frac{77\pm 13}{60}$ = $\frac{3}{2},\frac{16}{15}$

$m=\frac{3}{2},\frac{16}{15}$

Now, when m = $\frac{3}{2}$, $x-\frac{1}{x}=\frac{3}{2}$

$2x^2-3x-2=0$

$x=\frac{3\pm \sqrt{9+16}}{4}$

$x=\frac{3\pm 5}{4}$

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

Now, when $m=\frac{16}{15}$,

$x-\frac{1}{x}=\frac{16}{15}$

$15x^2-15=16x$
$15x^2-16x-15=0$

$x=\frac{16\pm \sqrt{256+900}}{30}$

$x=\frac{16\pm 34}{30}$

$x=\frac{5}{3},\frac{-3}{5}$

Hence, $x=2,\frac{-1}{2}.\frac{5}{3},\frac{-3}{5}$