SOLUTION: Find all real or imaginary solutions to each equation. Use the method of your choice. 3y^2 + 4y - 1 =0 thank you

Algebra ->  Exponents -> SOLUTION: Find all real or imaginary solutions to each equation. Use the method of your choice. 3y^2 + 4y - 1 =0 thank you      Log On


   

Question 177225: Find all real or imaginary solutions to each equation.
Use the method of your choice.
3y^2 + 4y - 1 =0
thank you
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
3y^2 + 4y - 1 =0
.
Your first attempt should be to try and factor the expression. In this case, you can't factor. Your next recourse is to use the quadratic equation. Doing so will yield the following two real solutions:
x = {0.215, -1.549}
.
Details of the quadratic follows:
.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B4x%2B-1+=+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=%284%29%5E2-4%2A3%2A-1=28.

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

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+28+%29%29%2F2%5C3+=+0.21525043702153
x%5B2%5D+=+%28-%284%29-sqrt%28+28+%29%29%2F2%5C3+=+-1.54858377035486

Quadratic expression 3x%5E2%2B4x%2B-1 can be factored:
3x%5E2%2B4x%2B-1+=+3%28x-0.21525043702153%29%2A%28x--1.54858377035486%29
Again, the answer is: 0.21525043702153, -1.54858377035486. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B4%2Ax%2B-1+%29