SOLUTION: Solve the following with substitution {{{ x^2-3x-sqrt(x^2-3x)=2 }}} My steps so far are {{{ U =sqrt(x^2-3x) }}} {{{ U^2 - U = 2 }}} {{{ U^2-U-2=0 }}} {{{ (U-2)(U+1) }

Algebra ->  Radicals -> SOLUTION: Solve the following with substitution {{{ x^2-3x-sqrt(x^2-3x)=2 }}} My steps so far are {{{ U =sqrt(x^2-3x) }}} {{{ U^2 - U = 2 }}} {{{ U^2-U-2=0 }}} {{{ (U-2)(U+1) }      Log On


   

Question 1094270: Solve the following with substitution
+x%5E2-3x-sqrt%28x%5E2-3x%29=2+
My steps so far are
+U+=sqrt%28x%5E2-3x%29+
+U%5E2+-+U+=+2+
+U%5E2-U-2=0+
+%28U-2%29%28U%2B1%29+
U= 2, -1
Then I went back using 2, -1 to plug back in
+U+=sqrt%28x%5E2-3x%29+
And ended up getting +sqrt%28-2%29+ and +sqrt%284%29+ from both
But -2 cant be it, so that would leave me with the 4 or 2,-2?
I'm not sure how to go about this.
Found 2 solutions by greenestamps, htmentor:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!

You used substitution to find U = sqrt(x^2-3x) is equal to either -1 or 2.

sqrt%28x%5E2-3x%29+=+-1 won't give you any solutions, because a square root is never negative.

sqrt%28x%5E2-3x%29+=+2
gives you solutions.
x%5E2-3x+=+4
x%5E2-3x-4+=+0
%28x-4%29%28x%2B1%29+=+0

The solutions are x=4 and x=-1; checking shows that both solutions work in the original equation.

Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
You made the incorrect substitution. You found that u = 2, -1,
but you substituted these values for x instead of u in the expression +U+=sqrt%28x%5E2-3x%29+
And we can reject u = -1 as a solution since a square root cannot be negative.
2 = sqrt%28x%5E2-3x%29+ -> x^2 - 3x - 4 = 0 -> (x-4)(x+1) = 0
So the two solutions are x = -1, x = 4