SOLUTION: find the roots of quadratic equations by factoring, x^2-6x+8

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Question 431617: find the roots of quadratic equations by factoring, x^2-6x+8
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
If x2-6x+8=0, then
x2-6x+8=0
(x-4)(x-2)=0
The roots of your function are 4 and 2.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%2B8+=+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=%28-6%29%5E2-4%2A1%2A8=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+4
x%5B2%5D+=+%28-%28-6%29-sqrt%28+4+%29%29%2F2%5C1+=+2

Quadratic expression 1x%5E2%2B-6x%2B8 can be factored:
1x%5E2%2B-6x%2B8+=+1%28x-4%29%2A%28x-2%29
Again, the answer is: 4, 2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B8+%29