Question

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

Answer 1

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 >= 0` and `8-x >=0`

`x >= -6` and `8 >=x`

Combining them:

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