SOLUTION: 25x^4-25x^2+6=0 i worked the problem out and got x=square root of 3/5 and 2/5 am i doing this right

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Question 443190: 25x^4-25x^2+6=0
i worked the problem out and got x=square root of 3/5 and 2/5 am i doing this right
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 25x%5E2%2B-25x%2B6+=+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=%28-25%29%5E2-4%2A25%2A6=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--25%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-25%29%2Bsqrt%28+25+%29%29%2F2%5C25+=+0.6
x%5B2%5D+=+%28-%28-25%29-sqrt%28+25+%29%29%2F2%5C25+=+0.4

Quadratic expression 25x%5E2%2B-25x%2B6 can be factored:
25x%5E2%2B-25x%2B6+=+25%28x-0.6%29%2A%28x-0.4%29
Again, the answer is: 0.6, 0.4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+25%2Ax%5E2%2B-25%2Ax%2B6+%29


It is not the square root of 2/5 and 3/5, but simply 2/5 and 3/5.