SOLUTION: i need a little help to make sure i am doing this right. Solve: 2x-5=(the square root of) x+20 this is what i have: (2x-5)^2=x+20 4x^2-10x-10x+25=x+20 4x^2-20x+25=x+20 4x

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: i need a little help to make sure i am doing this right. Solve: 2x-5=(the square root of) x+20 this is what i have: (2x-5)^2=x+20 4x^2-10x-10x+25=x+20 4x^2-20x+25=x+20 4x      Log On


   

Question 248126: i need a little help to make sure i am doing this right.
Solve: 2x-5=(the square root of) x+20
this is what i have:
(2x-5)^2=x+20
4x^2-10x-10x+25=x+20
4x^2-20x+25=x+20
4x^2-21x+5=0
(4x-1)(x-5)=0
4x-1=0 or x-5=0
x=1/4 or x=5
but the question on my review has multiple choice answers of:
a) 1/4 b)16 c) 29 or d) 5
which is the right answer or am i correct in thinking that there are 2 possible answers?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The solutions x=1%2F4 or x=5 satisfy the equation 4x%5E2-21x%2B5=0. However, we're not guaranteed that they satisfy the original equation 2x-5=sqrt%28x%2B20%29 (as extraneous roots may have been introduced).


To see if x=1%2F4 is a solution to 2x-5=sqrt%28x%2B20%29, simply plug it in to get:

2%281%2F4%29-5=sqrt%281%2F4%2B20%29


-9%2F2=sqrt%2881%2F4%29


-9%2F2=9%2F2 which is NOT true


So x=1%2F4 is NOT a solution of 2x-5=sqrt%28x%2B20%29


Similarly, plug in x=5 into 2x-5=sqrt%28x%2B20%29 to get:


2%285%29-5=sqrt%285%2B20%29


5=sqrt%2825%29


5=5


Since the equation is true, this verifies that x=5 is a solution.


So the only solution is x=5