The roots of frac{x+4}{x-4}+frac{x-4}{x+4}=fr | vedclass

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(C) Let $y = \frac{x+4}{x-4}$. Then the equation becomes $y + \frac{1}{y} = \frac{10}{3}$.Multiplying by $3y$,we get $3y^2 + 3 = 10y$,which simplifies

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$\pm 8$

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(C) Let $y = \frac{x+4}{x-4}$. Then the equation becomes $y + \frac{1}{y} = \frac{10}{3}$.Multiplying by $3y$,we get $3y^2 + 3 = 10y$,which simplifies to $3y^2 - 10y + 3 = 0$.Factoring the quadratic equation: $3y^2 - 9y - y + 3 = 0 \Rightarrow 3y(y-3) - 1(y-3) = 0 \Rightarrow (3y-1)(y-3) = 0$.Thus,$y = 3$ or $y = \frac{1}{3}$.Case $1$: $\frac{x+4}{x-4} = 3 \Rightarrow x+4 = 3x-12 \Rightarrow 2x = 16 \Rightarrow x = 8$.Case $2$: $\frac{x+4}{x-4} = \frac{1}{3} \Rightarrow 3x+12 = x-4 \Rightarrow 2x = -16 \Rightarrow x = -8$.Therefore,the roots are $x = 8, -8$ or $\pm 8$.

The roots of $\frac{x+4}{x-4}+\frac{x-4}{x+4}=\frac{10}{3}$ are:

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