Question

If 2( x² + 1 ) = 5x, find :
(i) ${\displaystyle x-\frac{1}{x}}$

(ii) ${\displaystyle x^3-\frac{1}{x^3}}$

Answer 1

(i) 2( x² + 1 ) = 5x
( x² + 1 ) = ${\displaystyle \frac{5}{2}}$x

Dividing by x, we have
${\displaystyle \frac{x^2+1}{x}=\frac{5}{2}}$

⇒ ${\displaystyle (x+\frac{1}{x})=\frac{5}{2}}$                           .....(1)
Now consider the expansion of  ${\displaystyle {(x+\frac{1}{x})}^2}$ :
${\displaystyle {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2}$

⇒ ${\displaystyle {\left(\frac{5}{2}\right)}^2=x^2+\frac{1}{x^2}+2}$                 [From(1)]

⇒ ${\displaystyle {\left(\frac{5}{2}\right)}^2-2=x^2+\frac{1}{x^2}}$

⇒ ${\displaystyle \frac{25}{4}-2=x^2+\frac{1}{x^2}}$

⇒ ${\displaystyle x^2+\frac{1}{x^2}=\frac{25-8}{4}}$

⇒ ${\displaystyle x^2+\frac{1}{x^2}=\frac{17}{4}}$                     ....(2)
Now consider the expansion of ${\displaystyle {(x-\frac{1}{x})}^2}$ :
${\displaystyle {(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\frac{17}{4}-2}$                 [from(2)]

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\frac{17-8}{4}}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\frac{9}{4}}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=\pm \frac{3}{2}}$                  ....(3)

(ii) We know that,
${\displaystyle (x^3-\frac{1}{x^3})={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})}$

∴ ${\displaystyle (x^3-\frac{1}{x^3})={(\pm \frac{3}{2})}^3+3(\pm \frac{3}{2})}$         [from(3)]

                             = ${\displaystyle \pm \frac{27}{8}+\frac{9}{2}}$

⇒ ${\displaystyle (x^3-\frac{1}{x^3})=\pm \frac{27+36}{8}}$

⇒ ${\displaystyle (x^3-\frac{1}{x^3})=\pm \frac{63}{8}}$