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

1 Answer
Aug 19, 2015

Answer:

#x = 3#

Explanation:

Since you're dealing with a radical term, recognize that you need to have

#x + 1 >=0" "# and #" "x-1>=0#

This is the case because, for real numbers, you can only take the square root of a positive number; moreover, the result of this operation will always be a positive number.

If #x-1>=0#, then you automatically have #x+1 >=0#, so you can combine these two conditions

#x -1 >= 0 implies x >=1#

Square both sides of the equation to get rid of the radical term

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

#x+color(red)(cancel(color(black)(1))) = x^2 - 2x + color(red)(cancel(color(black)(1)))#

This is equivalent to

#x^2 - 3x = 0#

#x(x - 3) = 0#

This equation can be euqal to zero if either #x=0# or #x-3=0#, which would imply

#x-3 = 0 implies x = 3#

Since #x=0# does not satisfy the initial contidion #x>=1#, it will be an extraneous solution. The only valid solution to your original equation will thus be #x = color(green)(3)#.

Do a quick check to make sure that you got the calculations right

#sqrt(3 + 1) = 3- 1#

#sqrt(4) = 2#

#2 = 2" "color(green)(sqrt())#