Question

${x}^{2} + \frac{1}{ \ {x}^{2} } + 2 - 3x - \frac{3}{2}$
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Answer 1

Answer:

$\dfrac{3 \pm \sqrt{5} }{2}$

Step-by-step explanation:

$x + \dfrac1x=t$

$x^2 +\dfrac{1}{x^2} +2=t^2$

Assuming the question is

$x^2 +\dfrac{1}{x^2} +2 -3x - \dfrac3x = 0$

$t^2 - 3t = 0$

$t(t-3) = 0$

$t = 0 \text{ OR } 3$

But t = 0 gives no real solutions, so if x is real, t = 3

$x + \dfrac1x = 3$

$x^2-3x+1=0$

$x = \dfrac{3 \pm \sqrt{9-4} }{2} = \dfrac{3 \pm \sqrt{5} }{2}$