SOLUTION: solvefor the zeros or roots of this polynomial equation q(x)=1-5x+6x^2 Please Please explain step by step because I really want to learn how to solve for the zeros of polynom

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: solvefor the zeros or roots of this polynomial equation q(x)=1-5x+6x^2 Please Please explain step by step because I really want to learn how to solve for the zeros of polynom      Log On


   

Question 164564: solvefor the zeros or roots of this polynomial equation
q(x)=1-5x+6x^2
Please Please explain step by step because I really want to learn how to solve for the zeros of polynomial equations. Thank you for any help. Please Help Soon!
Answer by KnightOwlTutor(293) About Me  (Show Source):
You can put this solution on YOUR website!
Use the quadratic formula.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 6x%5E2%2B-5x%2B1+=+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=%28-5%29%5E2-4%2A6%2A1=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+1+%29%29%2F2%5C6+=+0.5
x%5B2%5D+=+%28-%28-5%29-sqrt%28+1+%29%29%2F2%5C6+=+0.333333333333333

Quadratic expression 6x%5E2%2B-5x%2B1 can be factored:
6x%5E2%2B-5x%2B1+=+6%28x-0.5%29%2A%28x-0.333333333333333%29
Again, the answer is: 0.5, 0.333333333333333. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+6%2Ax%5E2%2B-5%2Ax%2B1+%29