SOLUTION: Find the real solutions of each equation. Please help me solve this equation {{{sqrt(2x+3)}}} - {{{sqrt(x+1)}}} = 1 Here's the steps I took: 1. {{{sqrt(2x+3)}}} = {{{sqrt(x+

Algebra ->  Radicals -> SOLUTION: Find the real solutions of each equation. Please help me solve this equation {{{sqrt(2x+3)}}} - {{{sqrt(x+1)}}} = 1 Here's the steps I took: 1. {{{sqrt(2x+3)}}} = {{{sqrt(x+      Log On


   

Question 181271This question is from textbook College Algebra Enhanced with Graphing Utilities, 5e
: Find the real solutions of each equation. Please help me solve this equation
sqrt%282x%2B3%29 - sqrt%28x%2B1%29 = 1
Here's the steps I took:
1. sqrt%282x%2B3%29 = sqrt%28x%2B1%2B1%29
2. 2x+3=x+1+1+1 sqrt%28x%2B1%2B1%29
3. 2x+3= x+2+1 sqrt%28x%2B1%29
4. x-1=1sqrt%28x%2B1%29
5. (x-1)^2=1(x+1)
6. x^2-2x+1=x+1
7. x^2-3x=0
8.(x-1)(x-2)=0
9. x=1, x=2
The answer in the book is -1 and 3 This question is from textbook College Algebra Enhanced with Graphing Utilities, 5e

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x%2B3%29-sqrt%28x%2B1%29=1
Your first step was incorrect.
You absorbed the 1 on the right hand side under the radical but you can't do that.
sqrt%282x%2B3%29-sqrt%28x%2B1%29=1
sqrt%282x%2B3%29=1%2Bsqrt%28x%2B1%29
%282x%2B3%29=%281%2Bsqrt%28x%2B1%29%29%5E2
2x%2B3=1%2B2%2Asqrt%28x%2B1%29%2B%28x%2B1%29
x%2B1=2%2Asqrt%28x%2B1%29
%28x%2B1%29%5E2=4%2A%28x%2B1%29
x%5E2%2B2x%2B1=4x%2B4
x%5E2-2x-3=0
%28x-3%29%28x%2B1%29=0
x=3, x=-1