How do you solve and check for extraneous solutions in #sqrt(x+7) = x + 1#?

1 Answer
Aug 10, 2015

Answer:

#x = 2#

Explanation:

Any solution that you will find must satisfy two conditions

  • #x+7>=0 implies x>= -7#
  • #x+1>=0 implies x>=-1#

Overall, the solution(s) to this equation must satisfy the condition #x>=-1#.

Start by squaring both sides of the equation to get rid of the radical term

#(sqrt(x+7))^2 = (x+1)^2#

#x+7 = x^2 + 2x + 1#

Rearrange this equation into classic quadratic form

#x^2 + x -6 = 0#

Use the quadratic formula to find the two solutions to this equation

#x_(1,2) = (-1 +- sqrt(1^2 - 4 * 1 * (-6)))/(2 * 1)#

#x_(1,2) = (-1 +- sqrt(25))/2#

#x_(1,2) = (-1 +- 5)/2 = {(x_1 = (-1 -5)/2 = -3), (x_2 = (-1 + 5)/2 = 2) :}#

Only one of these two solutions, #x=2#, satisfies the condition #x>=-1#, which means that #x=-3# will be an extraneous solution.

Therefore, the solution to this equation is #x=color(green)(2)#.

Check to see if this is the case

#sqrt(2 + 7) = 2 + 1#

#sqrt(9) = 3#

#3 = 3# #color(green)(sqrt())#