Given: #sqrt(5x + 20) - 7 = 2x - 4#
Isolate the square root function (radical) by adding #7# to both sides:
#sqrt(5x + 20) = 2x - 4 + 7#
#sqrt(5x + 20) = 2x +3#
Square both sides:
#5x + 20 = (2x+3)^2#
Use #(a + b)^2 = a^2 + 2ab + b^2# to distribute:
#5x + 20 = 4x^2 + 12x + 9#
Subtract #5x and 20# from both sides:
#0 = 4x^2 +7x -11#
Factor to solve:
#(4x + 11)(x - 1) = 0#
#4x + 11 = 0; " " x - 1 - 0#
#4x = -11; " " x = 1#
#x = -11/4#
You always need to check the solutions in the original equation to see if the are actually a valid solution.
CHECK: #x = 1#
#sqrt(5*1 + 20) - 7 = 2*1 - 4#
#sqrt(25) - 7 = 2 - 4#
#5 - 7 = -2#
#-2 = -2# TRUE #x = 1# is a valid solution
CHECK: #x = -11/4#
#sqrt(5*-11/4 + 20) - 7 = 2*-11/4 - 4#
#sqrt(-55/4 + 20) - 7 = -22/4 - 4#
#sqrt(-13.75 + 20) - 7 = -5.5 - 4#
#sqrt(6.25) - 7 = -9.5#
#2.5 - 7 = -9.5#
#-4.5 != -9.5# FALSE, so #x = -11/4# is not a valid solution
