Question

How do you solve `4=x+2sqrt(x-7)` and identify any restrictions?

Answer 1

`x = 6 +2i sqrt2, 6 -2i sqrt2`
There is no real root

Explanation:

`4 = x + 2 sqrt(x-7)`
`4 -x = 2 sqrt(x-7)`

square both sides
`(4 -x)^2 = (2 sqrt(x-7))^2`

`16 - 8 x +x^2 = 4(x-7)`
`16 - 8 x +x^2 = 4x- 28`

rearrange the equation,
`x^2 - 8x - 4x + 16 + 28 = 0`
`x^2 - 12x + 44 = 0`

`a = 1, b = -12, c =44`
since `b^2 - 4ac < 0`, so there is no real root for this equation.

We use completing a square to solve it.
`x^2 - 12x + 44 = 0`
`(x -6)^2 - (-6)^2 + 44 = 0`
`(x -6)^2 - 36 + 44 = 0`
`(x -6)^2 + 8 = 0`
`(x -6)^2 = -8`
`(x -6) = +-sqrt (-8) = +-sqrt (-1 * 8) = +- i sqrt 8 = +-2i sqrt2`
`x = 6 +-2i sqrt2`