How do you solve #sqrt(x + 2)+4 = x#?

1 Answer
Aug 5, 2015

Answer:

#x = 7#

Explanation:

Start by isolating the radical on one side of the equation by adding #-4# to both sides

#sqrt(x+2) + color(red)(cancel(color(black)(4))) - color(red)(cancel(color(black)(4))) = x-4#

#sqrt(x+2) = x-4#

This means that, in order to be considered a valid solution, the value(s) of #x# must satisfy the condition

#x-4>=0 <=> x>=4#

Next, square both sides of the equation to get rid of the square root

#(sqrt(x+2))^2 = (x-4)^2#

#x+2 = x^2 - 8x + 16#

Rearrange this equation to get all the terms on one side

#x^2 - 9x + 14 = 0#

You can find the two solutions to this quadratic by using the quadratic formula

#x_(1,2) = (-(-9) +- sqrt( (-9)^2 - 4 * 1 * (-14)))/(2 * 1)#

#x_(1,2) = (9 + sqrt(25))/2 = (9 +-5)/2 = {(x_1 = (9+5)/2 = color(green)(7)), (x_2 = (9-5)/2 = cancel(color(red)(2))) :}#

The only valid solution will be #x=7#; #x=2# will be an extraneous solution, since it does not satisfy the aforementioned condition #x>=4#.