Question

(x+1/x)^2-3/2(x-1/x)=4​

Answer 1

Step-by-step explanation:

$(x + \frac{1}{x} )^{2} - \frac{3}{2}(x - \frac{1}{x}) = 4 = > {x}^{2} + \frac{1}{ {x}^{2} }$

$+ 2.x. \frac{1}{x} - \frac{3}{2} (x - \frac{1}{x}) - 4 = 0 = >$

${x}^{2} + \frac{1}{ {x}^{2}} - 2 - \frac{3}{2} (x - \frac{1}{x} ) = 0 = >$

${(x - \frac{1}{x}) }^{2} - \frac{3}{2}( x - \frac{1}{x}) = 0 = >$

Let x -1/x be y Then

${y}^{2} - \frac{3}{2} y = 0 = > y(y - \frac{3}{2}) = 0 = >$

$y = 0 \: or \: \frac{3}{2}$

$if \: y = 0 = > x - \frac{1}{x} = 0 = > {x}^{2} - 1 =$

$0 = > (x - 1)(x + 1) = 0 = >$

$x = 1 \: or \: - 1$

$if \: y = \frac{3}{2} = > x - \frac{1}{x} = \frac{3}{2} = > {x}^{2} - \frac{3}{2}x$

$+ 1 = 0 = > {x}^{2} - 2x + \frac{x}{2} - 1 = 0 = >$

$x(x - 2) + \frac{1}{2} (x - 2) = 0 = >$

$(x + \frac{1}{2})(x - 2) = 0 = > x = 2\: or \: - \frac{1}{2}$

So the solutions of x which satisfy the above equation are 1,-1,2,-1/2. Hope it helps you.