SOLUTION: Alright, my past questions indicate that the equation is a rational equation. It's actually a radiacal equation. So heres the question: Find all real solutions in this equation:

Algebra ->  Rational-functions -> SOLUTION: Alright, my past questions indicate that the equation is a rational equation. It's actually a radiacal equation. So heres the question: Find all real solutions in this equation:      Log On


   

Question 661045: Alright, my past questions indicate that the equation is a rational equation. It's actually a radiacal equation. So heres the question:
Find all real solutions in this equation:
√(x-3)+1=√(x+2)
Please answer immediately, I really need your help since I need it tomorrow! Thanks!
Found 2 solutions by MathLover1, ewatrrr:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

sqrt%28x-3%29%2B1=sqrt%28x%2B2%29 ...raise both sides to power of 2

%28sqrt%28x-3%29%2B1%29%5E2=%28sqrt%28x%2B2%29%29%5E2

%28sqrt%28x-3%29%29%5E2%2B2sqrt%28x-3%29%2B1=%28sqrt%28x%2B2%29%29%5E2
x-3%2B2sqrt%28x-3%29%2B1=x%2B2
2sqrt%28x-3%29=x-x%2B3-1%2B2
2sqrt%28x-3%29=cross%28x%29-cross%28x%29%2B4
2sqrt%28x-3%29=4...raise both sides to power of 2
%282sqrt%28x-3%29%29%5E2=4%5E4
4%28x-3%29=16 .. both sides divide by 4
4%2F4%28x-3%29=16%2F4
x-3=4
x=4%2B3

highlight%28x=7%29

check:
sqrt%28x-3%29%2B1=sqrt%28x%2B2%29

sqrt%287-3%29%2B1=sqrt%287%2B2%29

sqrt%284%29%2B1=sqrt%289%29

2%2B1=3
3=3

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

√(x-3)+1=√(x+2)  || Squaring boths sides of the EQ

 (√(x-3)+1)(√(x-3)+1) = x+2

  x-3 + 2√(x-3)+ 1 = x+2    

    2√(x-3) = 4

     √(x-3) = 2  || Squaring boths sides of the EQ

       x-3 = 4

         x = 7