Question

(x+1/x)ˆ2 - 3/2(x-1/x)=4. Solve for x but x is not equal to 0

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{2} - \dfrac{3}{2}{\bigg(x - \dfrac{1}{x} \bigg) } = 4$

We know,

$\boxed{ \bf{ \: {(x + y)}^{2} = {(x - y)}^{2} + 4xy}}$

So, using this identity, the above equation can be rewritten as

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} + 4 \times x \times \dfrac{1}{x} - \dfrac{3}{2}{\bigg(x - \dfrac{1}{x} \bigg) } = 4$

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} + 4 - \dfrac{3}{2}{\bigg(x - \dfrac{1}{x} \bigg) } = 4$

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} + \cancel4 - \dfrac{3}{2}{\bigg(x - \dfrac{1}{x} \bigg) } = \cancel4$

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} - \dfrac{3}{2}{\bigg(x - \dfrac{1}{x} \bigg) } = 0$

$\rm :\longmapsto\:{\bigg(x - \dfrac{1}{x} \bigg) } \bigg \{{\bigg(x - \dfrac{1}{x} \bigg) } - \dfrac{3}{2} \bigg\} = 0$

Two cases arises

Case :- 1

$\rm :\longmapsto\:When \: \: \: x - \dfrac{1}{x}= 0$

$\rm :\longmapsto\:x = \dfrac{1}{x}$

$\rm :\longmapsto\: {x}^{2} = 1$

$\bf\implies \:x = \: \pm \: 1$

Case :- 2

$\rm :\longmapsto\:When \: \: \: x - \dfrac{1}{x} - \dfrac{3}{2} = 0$

$\rm :\longmapsto\:\: \: x - \dfrac{1}{x} = \dfrac{3}{2}$

$\rm :\longmapsto\:\: \: \dfrac{ {x}^{2} - 1}{x} = \dfrac{3}{2}$

$\rm :\longmapsto\: {2x}^{2} - 2 = 3x$

$\rm :\longmapsto\: {2x}^{2} - 3x - 2 = 0$

$\rm :\longmapsto\: {2x}^{2} - 4x + x - 2 = 0$

$\rm :\longmapsto\: {2x}(x - 2) +1( x - 2)= 0$

$\rm :\longmapsto\:(x - 2)(2x + 1) = 0$

$\bf\implies \:x = 2 \: \: or \: \: - \: \dfrac{1}{2}$

Hence,

The solution of equation

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{2} - \dfrac{3}{2}{\bigg(x - \dfrac{1}{x} \bigg) } = 4$

is

$\boxed{ \bf{ \: x \: = \: - 1, \: - \: \dfrac{1}{2}, \: 1, \: 2}}$