SOLUTION: Solve sqrt(x + 1)) = 1 - sqrt(2x) I started with sqrt of x + 1 = 1 - sqrt 2x^2. Next I got [1 - sqrt 2x]^2 = 2x - 2 sqrt 2x + 1. Not sure if I'm going the right route... An

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve sqrt(x + 1)) = 1 - sqrt(2x) I started with sqrt of x + 1 = 1 - sqrt 2x^2. Next I got [1 - sqrt 2x]^2 = 2x - 2 sqrt 2x + 1. Not sure if I'm going the right route... An      Log On


   

Question 36741: Solve sqrt(x + 1)) = 1 - sqrt(2x)

I started with sqrt of x + 1 = 1 - sqrt 2x^2. Next I got [1 - sqrt 2x]^2 = 2x - 2 sqrt 2x + 1. Not sure if I'm going the right route... Any help is appreciated
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%28x%2B1%29+=+1+-+sqrt%282x%29+

The aim is to get the square roots on one side so that we can square both sides to remove the square roots. As i say, that is the aim but in this question, we have to do some further work:

+sqrt%28x%2B1%29+%2B+sqrt%282x%29+=+1+
+%28sqrt%28x%2B1%29+%2B+sqrt%282x%29%29%5E2+=+1%5E2+
+%28sqrt%28x%2B1%29+%2B+sqrt%282x%29%29%2A%28sqrt%28x%2B1%29+%2B+sqrt%282x%29%29+=+1+

+%28x%2B1%29+%2B+sqrt%28x%2B1%29sqrt%282x%29+%2B+sqrt%282x%29sqrt%28x%2B1%29+%2B+2x+=+1+
+%28x%2B1%29+%2B+2sqrt%28x%2B1%29sqrt%282x%29+%2B+2x+=+1+
+3x%2B1+%2B+2sqrt%28x%2B1%29sqrt%282x%29+=+1+
+3x+%2B+2sqrt%28x%2B1%29sqrt%282x%29+=+0+
+3x+%2B+2sqrt%28%28x%2B1%29%2A%282x%29%29+=+0+
+2sqrt%28%28x%2B1%29%2A%282x%29%29+=+-3x+
+sqrt%28%28x%2B1%29%2A%282x%29%29+=+-3x%2F2+

and now square both sides again:
+%28sqrt%28%28x%2B1%29%2A%282x%29%29%29%5E2+=+%28-3x%2F2%29%5E2+
+%28%28x%2B1%29%2A%282x%29%29+=+9x%5E2%2F4+
+4%28%28x%2B1%29%2A%282x%29%29+=+9x%5E2+
+4%282x%5E2%2B2x%29+=+9x%5E2+
+8x%5E2%2B8x+=+9x%5E2+
+0+=+x%5E2+-+8x+
+x%5E2+-+8x+=+0+
x(x-8) = 0
so x=0 or x-8=0
--> x=0 or x=8

jon.