SOLUTION: I am stumped on this one. I have to Find all real solutions based on a fractional exponent? 2x-5 {{{ sqrt ( x ) }}} +2=0 The book wants me to solve the problem using substit

Question 223757: I am stumped on this one. I have to Find all real solutions based on a fractional exponent?
2x-5 +2=0
The book wants me to solve the problem using substitution. However, I can't find an equation that would make sense to me.
I am letting = 5 and w = x
I then come up with -5 +2w+2=0
This is where I draw a blank. I can't find a quadratic equation that will make that statement above true to continue solving.
I am assuming that I am doing this wrong, but beating my head against the book for the last hour now has only produced a squishy sound behind my skull! lol
Thank you in advance for help.
Kevin
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
2x-5 +2=0
The book wants me to solve the problem using substitution. However, I can't find an equation that would make sense to me.
I am letting = 5 and w = x
I then come up with -5 +2w+2=0
---------------
Sub w for sqrt(x)
--> 2w^2 - 5w + 2 = 0
(2w - 1)*(w - 2) = 0
w = 2
w = 1/2
------------
x = w^2
--> x = 4, x = 1/4

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