Question

How do you solve `sqrt(2x+5) - sqrt(x-2) = 3`?

Answer 1

`x=2` and `x=38`

Explanation:

Square both sides

`(sqrt(2x+5)-sqrt(x-2))(sqrt(2x+5)-sqrt(x-2))=(3)^2`

Expanding the left side we will have

`(2x+5)-2sqrt(x-2)sqrt(2x+5)+(x-2)=9`

Rewrite the product of the radicals on the left hand side

`2x+5-2sqrt((x-2)(2x+5))+x-2=9`

Combine like terms and expand under the radical on the left

`3x+3-2sqrt(2x^2+x-10)=9`

Subtract `9` from both sides and add `2sqrt(2x^2+x-10)` to both sides

`3x-6=2sqrt(2x^2+x-10)`

Now square both sides

`(3x-6)(3x-6)=(2sqrt(2x^2+x-10))^2`

Expand

`9x^2-36x+36=4(2x^2+x-10)`

Distribute the `4` on the right side

`9x^2-36x+36=8x^2+4x-40`

Subtract `8x^2` from both sides

Subtract `4x` from both sides

Add `40` to both sides

`x^2-40x+76=0`

This will factor to

`(x-2)(x-38)=0`

So `x-2=0`

`x=2`

or `x-38=0`

`x=38`