SOLUTION: solve by factoring completely 2x^3-4x^2+x = 2 I first factored x out of the left side of the equation x(2x^2-4x+1) = 2 I then applied the quadratic equation 4 +- sq rt of

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: solve by factoring completely 2x^3-4x^2+x = 2 I first factored x out of the left side of the equation x(2x^2-4x+1) = 2 I then applied the quadratic equation 4 +- sq rt of      Log On


   

Question 465506: solve by factoring completely
2x^3-4x^2+x = 2
I first factored x out of the left side of the equation
x(2x^2-4x+1) = 2
I then applied the quadratic equation
4 +- sq rt of 8/4
My end answer is 1 +- 1/2 rad 2. But when I plug it into the equation, it doesn't work. Where did I go wrong?
Answer by robertb(4012) About Me  (Show Source):
You can put this solution on YOUR website!
1) The equation is NOT in STANDARD FORM, that is, one side of the equation is not equal to zero, (so can't use the zero-product property just yet), and
2) Use the quadratic formula only for quadratic equations or for equations that are transformable to a quadratic equation.
2x%5E3-4x%5E2%2Bx+=+2 <==> 2x%5E3-4x%5E2%2Bx+-+2=0
<==> 2x%5E2%28x+-+2%29+%2B+1%2A%28x-2%29+=+0
<==> %282x%5E2+%2B+1%29%28x-2%29+=+0
==> x = 2 (from the factor x-2).
(There are NO REAL solutions from the factor 2x%5E2+%2B+1.)