How do you solve #(sqrt(x+5)-6)^2+4=20#?

Question

2032 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-14T09:45:28

The solution is #S={-1, 95}#

First, what's under the #sqrt # # "sign"# is #>=0#

Therefore,

#x+5>=0#

#x>=-5#

#(sqrt(x +5) - 6)^2 + 4 = 20#

#(sqrt(x +5) - 6)^2 = 16# //subtract both sides by 4

#sqrt(x +5) - 6 = ± 4# //take the square root of both sides

#sqrt(x +5) = 6± 4# //add 6 to both sides

#x +5 = (6± 4)^2# //square both sides

#x = (6± 4)^2-5# // subtract both sides by 5

so...

#x = (6± 4)^2-5#

#x_1 = (6+4)^2-5#
#x_1 = (10)^2-5#
#x_1=95#

#x_2 = (6-4)^2-5#
#x_2 = (2)^2-5#
#x_2 = -1#

The solution is #S={-1, 95}#

graph{(sqrt(x+5)-6)^2-16 [-84.1, 126.9, -46.6, 58.8]}