Question

How do you solve `\frac { x ^ { 2} + 64} { x ^ { 2} - 64} = \frac { x } { x + 8} - \frac { 8} { x - 8}`?

Answer 1

there are no solutions to the given equation

Explanation:

We have:

`(x^2+64)/(x^2-64) = x/(x+8)-8/(x-8)`

`:. (x^2+64)/(x^2-64) = (x(x-8)-8(x+8)) / ( (x+8)(x-8) )`

`:. (x^2+64)/(x^2-64) = { x^2-8x-8x-64 } / ( x^2-8x+8x-64 )`

`:. (x^2+64)/(x^2-64) = { x^2 - 16x -64 } / ( x^2-64 )`

The denominator is zero, if

`x^2-64 = 0 => x^2=64 => x = +- 8`

Providing that the denominator is nonzero, then we have:

`x^2+64 = x^2 - 16x -64`
`:. 16x = -128`
`:. x = -8`

However because this value would make the denominator zero it has to be excluded, hence there are no solutions to the given equation