SOLUTION: I thought I new how to do these but I guess I was wrong! Please help me! Solve (radical)(x-3) + (radical)(x-2) = 5 I know you have to isolate the radicals on opposite s

Question 61230This question is from textbook College Algebra Graphs and Models
: I thought I new how to do these but I guess I was wrong! Please help me!
Solve
(radical)(x-3) + (radical)(x-2) = 5
I know you have to isolate the radicals on opposite sides of the equation, but I have no idea how to solve after that! Please help!
Thank you! This question is from textbook College Algebra Graphs and Models

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
(radical)(x-3) + (radical)(x-2) = 5
:
SqRt(x-3) = 5 - SqRt(x-2)
:
Square both sides, FOIL (5-SqRt(x-2))^2 and you have:
(x - 3) = 25 - 10*SqRt(x-2) + (x-2)
:
x - 3 = 25 - 2 - 10*SqRt(x-2) + x
x - 3 = 23 - 10*SqRt(x-2) + x
:
Isolate the radical on the right:
x - x - 3 - 23 = -10*SqRt(x-2)
-26 = -10*SqRt(x-2)
:
Square both sides again:
+676 = +100(x-2)
+676 = 100x - 200
676 + 200 = 100x
876 = 100x
x = 876/100
x = 8.76:
:
:
Check using 8.76 for x in original equation;
SqRt(5.76) + SqRt(6.76) =
2.4 + 2.6 = 5
:
You got it now?

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