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anonymous
 5 years ago
I lied, I have one more :)
anonymous
 5 years ago
I lied, I have one more :)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y ^{4}3y ^{2}88=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Again, substitute something for y^2 to get a normal quadratic you can solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's just like the others you were doing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0i meant it the other way around y 2 for the y 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use u instead of y for the second variable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it'll be less confusing. Let \(u = y^2\) So \(?\ =y^4\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. So now in your original equation replace all the \(y^2\) with u and all the \(y^4\) with \(u^2\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the do i want to move everything else to the other side, where it's 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, you need the 0 there in order to use the quadratic equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[au^2 + bu + c = 0\] Means: \[u = \frac{(b) \pm \sqrt{b^24(a)(c)}}{2(a)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so plug in your a, b, and c and solve for u, what do you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0That's not the right square root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{(3)^2  4(1)(88)} = \sqrt{9 +352} = \sqrt{361} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what do you get with that correction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, I mean for the whole equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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
 5 years ago
Best ResponseYou've already chosen the best response.02a is not 19, it's 2. Look, a= 1, b= 3, c= 88 \[u = \frac{(3) \pm \sqrt{(3)^24(1)(88)}}{2(1)}\] \[ = \frac{3 \pm \sqrt{361}}{2} = \frac{3 \pm 19}{2} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what are the two values we get for u?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.011 and 11. 361 squared is 19.. so how are there 4 answers?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because those are the answers for u. 11, and 8. But \(u = y^2\) So that means y can be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I already typed the other one in and got it wrong. ha. oops. but thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, since u = y^2 u cannot be 8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0So there's only 2 solutions. u = \(\pm \sqrt{11}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep, thats what it said. thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have to go to bed I'm afraid. Good luck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0THanks for all your help! night
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