Question

How do you solve the quadratic `2/(x+5)-x/(x-5)=1` using any method?

Answer 1

Stipulate that `x !=+-5`. Multiply both sides by `(x - 5)(x + 5)`. Collect all the terms on one side. Factor or use the quadratic formula. Discard any restricted roots.

Explanation:

Stipulate that `x !=+-5`:

`2/(x + 5) - x/(x - 5) = 1; x !=+-5`

Multiply both sides by `(x - 5)(x + 5)`:

`2(x - 5) - x(x + 5) = x^2 - 25; x !=+-5`

`2x - 10 - x^2 - 5x = x^2 - 25; x !=+-5`

Collect all of the terms to one side:

`2x^2 + 3x - 15 = 0;x !=+-5`

Check the discriminant:

`d = b^2 -4(a)(c) = 3^2 - 4(2)(-15) = 129`

Drop the restriction, because, with 129 under the radical, there is no way to obtain a root equal to `+-5`.

Use the quadratic formula:

`x = (-b +-sqrt(d))/(2a)`

`x = (-3 +-sqrt(129))/(2(2))`

`x = {((-3 +sqrt(129))/(4)),((-3 -sqrt(129))/(4)):}`