SOLUTION: how do I solve this equation? I have to find the real roots for {{{y^(-2)-2/y +2=0}}}

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Question 182706This question is from textbook
: how do I solve this equation? I have to find the real roots for y%5E%28-2%29-2%2Fy+%2B2=0 This question is from textbook

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Multiply thru by y^2 to get
1+-+2y+%2B+2y%5E2+=+0
Now use quadratic equation to solve
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 2y%5E2%2B-2y%2B1+=+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=%28-2%29%5E2-4%2A2%2A1=-4.

The discriminant -4 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -4 is + or - sqrt%28+4%29+=+2.

The solution is y%5B12%5D+=+%28--2%2B-+i%2Asqrt%28+-4+%29%29%2F2%5C2+=++%28--2%2B-+i%2A2%29%2F2%5C2+

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-2%2Ax%2B1+%29