# How do you solve sqrt(8-x)=x+6 and identify any restrictions?

May 27, 2017

#### Answer:

The right side cannot be negative and neither can the argument of the radical.
Square both sides.
Solve the quadratic.
Discard any roots that violate the restrictions.

#### Explanation:

The restrictions are that the right side cannot be negative and the same for the argument of the radical:

$x + 6 \ge 0$ and $8 - x \ge 0$

$x \ge - 6$ and $8 \ge x$

Combining them:

$- 6 \le x \le 8$

Add this restriction to the given equation:

sqrt(8-x)=x+6; -6 <= x <= 8

Square both sides:

8-x = x^2+ 12x+ 36; -6 <= x <= 8

Combine like terms:

0 = x^2+ 13x+ 28; -6 <= x <= 8

Use the quadratic formula:

x = (-13+-sqrt(13^2-4(1)(28)))/(2(1)); -6 <= x <= 8

x = (-13+-sqrt(57))/2; -6 <= x <= 8

The negative root us outside the restriction, therefore, we discard it:

$x = \frac{- 13 + \sqrt{57}}{2}$