SOLUTION: can you please help me use the quadratic formula to solve the following and leave irrational roots in the simplest radical form: (((2x^2-3x+1=0)))

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Question 179730This question is from textbook algebra structure and method
: can you please help me use the quadratic formula to solve the following and leave irrational roots in the simplest radical form: (((2x^2-3x+1=0))) This question is from textbook algebra structure and method

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2-3x%2B1=0
:
%282x-1%29%28x-1%29=0
:
system%28x=1%2F2%2Cx=1%29
:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-3x%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=%28-3%29%5E2-4%2A2%2A1=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+1+%29%29%2F2%5C2+=+1
x%5B2%5D+=+%28-%28-3%29-sqrt%28+1+%29%29%2F2%5C2+=+0.5

Quadratic expression 2x%5E2%2B-3x%2B1 can be factored:
2x%5E2%2B-3x%2B1+=+%28x-1%29%2A%28x-0.5%29
Again, the answer is: 1, 0.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-3%2Ax%2B1+%29