SOLUTION: 5x^(2/5)-7x^(1/5)+2=0

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Question 253295: 5x^(2/5)-7x^(1/5)+2=0
Answer by JimboP1977(311) About Me  (Show Source):
You can put this solution on YOUR website!
5x%5E%282%2F5%29-7x%5E%281%2F5%29%2B2=0
5%28x%5E%281%2F5%29%29%5E2-7x%5E%281%2F5%29%2B2=0
5%28y%29%5E2-7y%2B2=0 Let y=x%5E%281%2F5%29
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 5y%5E2%2B-7y%2B2+=+0) has the following solutons:

y%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=%28-7%29%5E2-4%2A5%2A2=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--7%2B-sqrt%28+9+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%28-7%29%2Bsqrt%28+9+%29%29%2F2%5C5+=+1
y%5B2%5D+=+%28-%28-7%29-sqrt%28+9+%29%29%2F2%5C5+=+0.4

Quadratic expression 5y%5E2%2B-7y%2B2 can be factored:
5y%5E2%2B-7y%2B2+=+5%28y-1%29%2A%28y-0.4%29
Again, the answer is: 1, 0.4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B-7%2Ax%2B2+%29


1=x^(1/5), x=1
0.4=x^(1/5),x=0.010204