If x + frac{1}{x} = sqrt{13},then frac{3x}{x^ | vedclass

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Question

(C) Given that $x + \frac{1}{x} = \sqrt{13}$.Divide the numerator and denominator of the expression $\frac{3x}{x^2 - 1}$ by $x$:$\frac{3x/x}{(x^2 - 1)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given that $x + \frac{1}{x} = \sqrt{13}$.Divide the numerator and denominator of the expression $\frac{3x}{x^2 - 1}$ by $x$:$\frac{3x/x}{(x^2 - 1)/x} = \frac{3}{x - \frac{1}{x}}$.We know that $(x - \frac{1}{x})^2 = (x + \frac{1}{x})^2 - 4$.Substituting the given value: $(x - \frac{1}{x})^2 = (\sqrt{13})^2 - 4 = 13 - 4 = 9$.Therefore,$x - \frac{1}{x} = \pm 3$.Substituting this into our expression: $\frac{3}{\pm 3} = \pm 1$.Since the options provided are limited,we consider the magnitude or the positive case,which is $1$.

If $x + \frac{1}{x} = \sqrt{13}$,then $\frac{3x}{x^2 - 1}$ is equal to:

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