Question

(x+1/x)^2 = 2(x+1/x)+3 solve this equation. [DO NOT SEND ANY OTHER MESSAGE ]​

Answer 1

Step-by-step explanation:

Let (x+1/x) be y. So

${y}^{2} = 2y + 3 = > {y}^{2} - 2y - 3 = 0 = >$

${y}^{2} - 3y + y - 3 = 0 = >$

$y(y -3) + 1(y - 3) = 0 = >$

$(y + 1)(y - 3) = 0 = > y = - 1 \: or \: 3$

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

${x}^{2} + x + 1 = 0 = > x = \frac{ - 1 + \sqrt{1 - 4} }{2} \: or \:$

$\frac{- 1 - \sqrt{1 - 4}}{2}$

x = (-1+√-3)/2 and (-1-√-3)/2.

$x + \frac{1}{x} = 3 = > {x}^{2} + 1 = 3x = >$

${x}^{2} -3x + 1 = 0 = > x = \frac{3 + \sqrt{9 - 4} }{2} \: or \:$

$\frac{3 - \sqrt{9 - 4} }{2} = > x = \frac{3 + \sqrt{5} }{2} \: or \: \frac{3 - \sqrt{5} }{2}$

Hope it helps you.