Reduce the following equation into quadratic and solve.
${(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4$

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Solution

Put $x + \frac{1}{x} = t$
$\Rightarrow t^{2} - 3 t - 4 = 0$
$(t - 4) (t + 1) = 0$
$\Rightarrow t = 4, - 1$
$\Rightarrow x + \frac{1}{x} = 4$ $x + \frac{1}{x} = - 1$
$\Rightarrow x^{2} - 4 x + 1 = 0$ $x^{2} + x + 1 = 0$
$x = 2 \pm \sqrt{3}$ $x = \frac{- 1 \pm \sqrt{3} i}{2}$

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Reduce the following equation into quadratic and solve.(x+1x)2=3(x+1x)+4

Put x+1x=t⇒t2−3t−4=0(t−4)(t+1)=0⇒t=4,−1⇒x+1x=4 x+1x=−1⇒x2−4x+1=0 x2+x+1=0x=2±√3 x=−1±√3i2

Reduce the following equation into quadratic and solve.
${(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4$

Put $x + \frac{1}{x} = t$
$\Rightarrow t^{2} - 3 t - 4 = 0$
$(t - 4) (t + 1) = 0$
$\Rightarrow t = 4, - 1$
$\Rightarrow x + \frac{1}{x} = 4$ $x + \frac{1}{x} = - 1$
$\Rightarrow x^{2} - 4 x + 1 = 0$ $x^{2} + x + 1 = 0$
$x = 2 \pm \sqrt{3}$ $x = \frac{- 1 \pm \sqrt{3} i}{2}$