The value of x in the equation left(x+frac{1} | vedclass

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(C) Let $y = x - \frac{1}{x}$.Then,$y^2 = x^2 + \frac{1}{x^2} - 2$,so $x^2 + \frac{1}{x^2} = y^2 + 2$.We know that $(x + \frac{1}{x})^2 = x^2 + \frac{

Vedclass · Accepted Answer

$-1$

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(C) Let $y = x - \frac{1}{x}$.Then,$y^2 = x^2 + \frac{1}{x^2} - 2$,so $x^2 + \frac{1}{x^2} = y^2 + 2$.We know that $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$.Substituting the value,$(x + \frac{1}{x})^2 = (y^2 + 2) + 2 = y^2 + 4$.The given equation is $(x + \frac{1}{x})^2 - \frac{3}{2}(x - \frac{1}{x}) = 4$.Substituting $y$,we get $(y^2 + 4) - \frac{3}{2}y = 4$.$y^2 - \frac{3}{2}y = 0$.$y(y - \frac{3}{2}) = 0$.So,$y = 0$ or $y = \frac{3}{2}$.Case $1$: $x - \frac{1}{x} = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.Case $2$: $x - \frac{1}{x} = \frac{3}{2} \Rightarrow 2x^2 - 2 = 3x \Rightarrow 2x^2 - 3x - 2 = 0$.$(2x + 1)(x - 2) = 0 \Rightarrow x = -\frac{1}{2}$ or $x = 2$.The possible values for $x$ are $1, -1, 2, -\frac{1}{2}$. Among the given options,$-1$ is a valid solution.

The value of $x$ in the equation $\left(x+\frac{1}{x}\right)^{2}-\frac{3}{2}\left(x-\frac{1}{x}\right)=4$ is

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