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.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

hmm..

Give it a try

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?

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

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.

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?

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