SOLUTION: find the real number solutions of the equation. y^4-14y^2+45

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Question 175737: find the real number solutions of the equation.
y^4-14y^2+45
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Your first problem is that you don't actually have an equation. There is not an equal sign to be seen anywhere in y%5E4-14y%5E2%2B45. So all I can do is assume that you meant y%5E4+-+14y%5E2+%2B+45+=+0. Presuming this is the case, let t+=+y%5E2

Then you can write: t%5E2+-+14t+%2B+45+=+0 which is a quadratic that can be solved by ordinary means; this one factors because +%28-5%29+%2A+%28-9%29+=+45 and %28-5%29+%2B+%28-9%29+=+-14, so:

t%5E2+-+14t+%2B+45+=+%28t+-+5%29%28t+-+9%29+=+0

Therefore, t+-+5+=+0 or t+-+9+=+0 which is to say:

t+=+5 or t+=+9

But remember that t+=+y%5E2, so we can now say:

y%5E2+=+5 which means that y+=+sqrt%285%29 or y+=+-sqrt%285%29, and

y%5E2+=+9 which means that y+=+sqrt%289%29+=+3 or y+=+-sqrt%289%29+=+-3

And there are your four real number roots.