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

1 Answer
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 >= 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#