How do you simplify #sqrt(1+x) - sqrt(1-x)#?

1 Answer
Sep 29, 2015

Answer:

#sqrt(1+x) - sqrt(1-x)# does not really simplify,

but you can re-express it in various ways...

Explanation:

First note that for both square roots to have Real values, we must have #x in [-1, 1]#

Let's see what happens when you square #sqrt(1+x) - sqrt(1-x)# ...

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

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

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

[[ using #sqrt(a)sqrt(b) = sqrt(ab)# ]]

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

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

What is the correct sign to choose?

If #x >= 0# then #1 + x >= 1 - x#, so #sqrt(1+x)-sqrt(1-x) >= 0#

If #x < 0# then #1 + x < 1 - x#, so #sqrt(1+x)-sqrt(1-x) < 0#

So we have:

#sqrt(1+x) - sqrt(1-x) = { (sqrt(2 - 2sqrt(1-x^2)), "if " x >= 0), (-sqrt(2 - 2sqrt(1-x^2)), "if " x < 0) :}#

If you like, you can separate out the common factor #2# to get:

#sqrt(1+x) - sqrt(1-x) = { (sqrt(2)sqrt(1 - sqrt(1-x^2)), "if " x >= 0), (-sqrt(2)sqrt(1 - sqrt(1-x^2)), "if " x < 0) :}#

graph{sqrt(1+x)-sqrt(1-x) [-5, 5, -2.5, 2.5]}