SOLUTION: I was asked to find the square root of my equation which is 4x^2-8x+4=56. I know that I should move the 4 to the other side so my equation would then be 4x^2-8x=52, and I know that

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: I was asked to find the square root of my equation which is 4x^2-8x+4=56. I know that I should move the 4 to the other side so my equation would then be 4x^2-8x=52, and I know that      Log On


   

Question 481285: I was asked to find the square root of my equation which is 4x^2-8x+4=56. I know that I should move the 4 to the other side so my equation would then be 4x^2-8x=52, and I know that the square root of 4x^2 would be 2x and I know that the square root of 52 is 7.21, but what would be the square root of 8x? That is where I get lost.
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2-8x+4=56
4x2-8x-52=0
x2-2x-13=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-13+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A1%2A-13=56.

Discriminant d=56 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+56+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+56+%29%29%2F2%5C1+=+4.74165738677394
x%5B2%5D+=+%28-%28-2%29-sqrt%28+56+%29%29%2F2%5C1+=+-2.74165738677394

Quadratic expression 1x%5E2%2B-2x%2B-13 can be factored:
1x%5E2%2B-2x%2B-13+=+1%28x-4.74165738677394%29%2A%28x--2.74165738677394%29
Again, the answer is: 4.74165738677394, -2.74165738677394. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-13+%29