SOLUTION: This is my original question and I keep coming up with the same answer as the tutor but my book says the answers are -5 and 2. Is that possible? 2x^2-x=1 original equation 2x^2

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: This is my original question and I keep coming up with the same answer as the tutor but my book says the answers are -5 and 2. Is that possible? 2x^2-x=1 original equation 2x^2      Log On


   

Question 30689: This is my original question and I keep coming up with the same answer as the tutor but my book says the answers are -5 and 2. Is that possible?
2x^2-x=1 original equation
2x^2-x-1=0 set the equation to zero
2x^2-2x+x-1=0 the product leading coefficient and the constant is -2 (two numbers that add up to -1 and are multiplied to get the -2 is: -2 and 1)
(2x^2-2x)+(x-1)=0 start to combine
2x(x-1)+1(x-1)=0 take out the factors
(2x+1)(x-1)=0 set them to zero
2x+1 = 0 ; 2x = -1 ; x = -1/2
x-1 = 0 ; x = 1
1 and (-1/2) are your answers
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
2x%5E2-x+=+1 Subtract 1 from both sides of the equation.
2x%5E2-x-1+=+0 Factor.
%282x%2B1%29%28x-1%29+=+0 Apply the zero products principle.
2x%2B1+=+0 and/or x-1+=+0
If 2x%2B1+=+0 then 2x+=+-1 and x+=+-1%2F2
If x-1+=+0 then x+=+1
The roots are:
x+=+-1%2F2
x+=+1
If the roots were:
x+=+-5 and x+=+2 then the original equation has to be:
x%5E2%2B3x-10+=+0 because:
%28x%2B5%29%28x-2%29+=+x%5E2%2B3x-10
Check your book again. Math books have been known to be in error!