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

Mar 24, 2017

$x = 6 + 2 i \sqrt{2} , 6 - 2 i \sqrt{2}$
There is no real root

#### Explanation:

$4 = x + 2 \sqrt{x - 7}$
$4 - x = 2 \sqrt{x - 7}$

square both sides
${\left(4 - x\right)}^{2} = {\left(2 \sqrt{x - 7}\right)}^{2}$

$16 - 8 x + {x}^{2} = 4 \left(x - 7\right)$
$16 - 8 x + {x}^{2} = 4 x - 28$

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

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

We use completing a square to solve it.
${x}^{2} - 12 x + 44 = 0$
${\left(x - 6\right)}^{2} - {\left(- 6\right)}^{2} + 44 = 0$
${\left(x - 6\right)}^{2} - 36 + 44 = 0$
${\left(x - 6\right)}^{2} + 8 = 0$
${\left(x - 6\right)}^{2} = - 8$
$\left(x - 6\right) = \pm \sqrt{- 8} = \pm \sqrt{- 1 \cdot 8} = \pm i \sqrt{8} = \pm 2 i \sqrt{2}$
$x = 6 \pm 2 i \sqrt{2}$