Question

Solve:

${\displaystyle 9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0}$

Answer 1

Given equation be,

${\displaystyle 9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0}$

Put ${\displaystyle x+\frac{1}{x}=y}$

Then ${\displaystyle x^2+\frac{1}{x^2}+2=y^2}$  ...[On squaring]

${\displaystyle \Rightarrow x^2+\frac{1}{x^2}=y^2-2}$

Thus given equation

∴ ${\displaystyle 9(x^2+\frac{1}{x^2})-9(x+\frac{1}{x})-52=0}$

${\displaystyle \Rightarrow }$ 9(y² – 2) – 9y – 52 = 0

${\displaystyle \Rightarrow }$ 9y² – 18 – 9y – 52 = 0

${\displaystyle \Rightarrow }$ 9y² – 9y – 70 = 0

${\displaystyle \Rightarrow }$ 9y² – 30y + 21y – 70 = 0

${\displaystyle \Rightarrow }$ 3y(3y – 10) + 7(3y – 10) = 0

${\displaystyle \Rightarrow }$ (3y – 10)(3y + 7) = 0 

Either 3y – 10 = 0, then y = ${\displaystyle \frac{10}{3}}$

Or 3y + 7 = 0 then y = ${\displaystyle \frac{-7}{3}}$

(i) When y = ${\displaystyle \frac{10}{3}}$, then ${\displaystyle x+\frac{1}{x}=\frac{10}{3}}$

${\displaystyle \Rightarrow }$ 3x² + 3 = 10x

${\displaystyle \Rightarrow }$ 3x² – 10x + 3 = 0

${\displaystyle \Rightarrow }$ 3x² – 9x – x + 3 = 0

${\displaystyle \Rightarrow }$ 3x(x – 3) – 1(x – 3) = 0

${\displaystyle \Rightarrow }$ (x – 3)(3x – 1) = 0

Either x – 3 = 0, then x = 3

Or 3x – 1 = 0, then x = ${\displaystyle \frac{1}{3}}$

(ii) When ${\displaystyle y=\frac{-7}{3}}$, then ${\displaystyle x+\frac{1}{x}=\frac{-7}{3}}$

${\displaystyle \Rightarrow }$ 3x² + 3 = –7x

${\displaystyle \Rightarrow }$ 3x² + 7x + 3 = 0

Here a = 3, b = 7, c = 3

∴ D = b² – 4ac

= (7)² – 4 × 3 × 3

= 49 – 36

= 13

∴ ${\displaystyle x=\frac{-b\times \sqrt{D}}{2a}}$

= ${\displaystyle \frac{-7\pm \sqrt{13}}{2\times 3}}$

= ${\displaystyle \frac{-7\pm \sqrt{13}}{6}}$

∴ x = ${\displaystyle 3,\frac{1}{3},\frac{-7\pm \sqrt{13}}{6}}$