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

##### 1 Answer

#### Answer:

#### 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 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-3 = 0 implies x = 3#

Since *extraneous solution*. The only valid solution to your original equation will thus be

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

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

#sqrt(4) = 2#