Question

`\sqrt15-x +\sqrt3-x=6`

Answer 1

As currently written, solving for
`x`, we have

`√(15) - x + \sqrt3 - x = 6`

`2x = √(15) + \sqrt3 - 6`

`x = \frac{√(15) +\sqrt3 - 6}2`

I suspect you meant to write

`√(15 - x) + √(3 - x) = 6`

Move one of the roots to the other side, then take squares on both sides.

`√(15 - x) = 6 - √(3 - x)`

`\left(√(15 - x)\right)^2 = \left(6 - √(3 - x)\right)^2`

`15 - x = 36 - 12 √(3 - x) + (3 - x)`

`12 √(3 - x) = 24`

Take squares again

`\left(12√(3-x)\right)^2 = 24^2`

`144 (3 - x) = 576`

`432 - 144x = 576`

`144x = -144`

`\boxed{x = -1}`