I lied, I have one more :)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

I lied, I have one more :)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[y ^{4}-3y ^{2}-88=0\]
Again, substitute something for y^2 to get a normal quadratic you can solve
It's just like the others you were doing

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

hmm..
Give it a try
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.
i meant it the other way around y 2 for the y 4
Use u instead of y for the second variable
it'll be less confusing. Let \(u = y^2\) So \(?\ =y^4\)
u^2?
Yep. So now in your original equation replace all the \(y^2\) with u and all the \(y^4\) with \(u^2\)
the do i want to move everything else to the other side, where it's 0?
No, you need the 0 there in order to use the quadratic equation.
\[au^2 + bu + c = 0\] Means: \[u = \frac{-(b) \pm \sqrt{b^2-4(a)(c)}}{2(a)}\]
Look familiar?
yes it does!
Ok, so plug in your a, b, and c and solve for u, what do you get?
3+/- \[\sqrt{-167}\]
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?
That's not the right square root
\[\sqrt{(-3)^2 - 4(1)(-88)} = \sqrt{9 +352} = \sqrt{361} \]
So what do you get with that correction?
361
No, I mean for the whole equation.
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
2a is not 19, it's 2. Look, a= 1, b= -3, c= -88 \[u = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-88)}}{2(1)}\] \[ = \frac{3 \pm \sqrt{361}}{2} = \frac{3 \pm 19}{2} \]
So what are the two values we get for u?
11 and -11. 361 squared is 19.. so how are there 4 answers?
Because those are the answers for u. 11, and -8. But \(u = y^2\) So that means y can be?
I already typed the other one in and got it wrong. ha. oops. but thanks
Actually, since u = y^2 u cannot be -8
oh. limited tries?
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
So there's only 2 solutions. u = \(\pm \sqrt{11}\)
yep, thats what it said. thanks!
I have to go to bed I'm afraid. Good luck
THanks for all your help! night

Not the answer you are looking for?

Search for more explanations.

Ask your own question