I lied, I have one more :)

- anonymous

I lied, I have one more :)

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- anonymous

\[y ^{4}-3y ^{2}-88=0\]

- anonymous

Again, substitute something for y^2 to get a normal quadratic you can solve

- anonymous

It's just like the others you were doing

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## More answers

- anonymous

hmm..

- anonymous

Give it a try

- anonymous

i substituted y^4 for y^2 and the y^2 turned into just a y?.. ahh. Idk why this is so difficult for me. I'm sure it's simple.

- anonymous

i meant it the other way around y 2 for the y 4

- anonymous

Use u instead of y for the second variable

- anonymous

it'll be less confusing.
Let \(u = y^2\)
So \(?\ =y^4\)

- anonymous

u^2?

- anonymous

Yep. So now in your original equation replace all the \(y^2\) with u and all the \(y^4\) with \(u^2\)

- anonymous

the do i want to move everything else to the other side, where it's 0?

- anonymous

No, you need the 0 there in order to use the quadratic equation.

- anonymous

\[au^2 + bu + c = 0\]
Means:
\[u = \frac{-(b) \pm \sqrt{b^2-4(a)(c)}}{2(a)}\]

- anonymous

Look familiar?

- anonymous

yes it does!

- anonymous

Ok, so plug in your a, b, and c and solve for u, what do you get?

- anonymous

3+/- \[\sqrt{-167}\]

- anonymous

so if it's in the root it would actually be positive. i'm trying to remember.. do I incorporate the 3 with the -167 at all? or did I accurately solve this even?

- anonymous

That's not the right square root

- anonymous

\[\sqrt{(-3)^2 - 4(1)(-88)} = \sqrt{9 +352} = \sqrt{361} \]

- anonymous

So what do you get with that correction?

- anonymous

361

- anonymous

No, I mean for the whole equation.

- anonymous

It came down to 361=19 and those divided =19.. so would it just come to 19 or is there more numbers to that? Everything seemed to cancel out

- anonymous

2a is not 19, it's 2.
Look, a= 1, b= -3, c= -88
\[u = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-88)}}{2(1)}\]
\[ = \frac{3 \pm \sqrt{361}}{2} = \frac{3 \pm 19}{2} \]

- anonymous

So what are the two values we get for u?

- anonymous

11 and -11. 361 squared is 19.. so how are there 4 answers?

- anonymous

Because those are the answers for u.
11, and -8. But \(u = y^2\)
So that means y can be?

- anonymous

I already typed the other one in and got it wrong. ha. oops. but thanks

- anonymous

Actually, since u = y^2 u cannot be -8

- anonymous

oh. limited tries?

- anonymous

i have one more problem I have to get right. it's a different problem if you want me to type it on here. otherwise I'll post it to the left

- anonymous

So there's only 2 solutions.
u = \(\pm \sqrt{11}\)

- anonymous

yep, thats what it said. thanks!

- anonymous

I have to go to bed I'm afraid. Good luck

- anonymous

THanks for all your help! night

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