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

1. anonymous

$y ^{4}-3y ^{2}-88=0$

2. anonymous

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

3. anonymous

It's just like the others you were doing

4. anonymous

hmm..

5. anonymous

Give it a try

6. 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.

7. anonymous

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

8. anonymous

Use u instead of y for the second variable

9. anonymous

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

10. anonymous

u^2?

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

Look familiar?

16. anonymous

yes it does!

17. anonymous

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

18. anonymous

3+/- $\sqrt{-167}$

19. 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?

20. anonymous

That's not the right square root

21. anonymous

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

22. anonymous

So what do you get with that correction?

23. anonymous

361

24. anonymous

No, I mean for the whole equation.

25. 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

26. anonymous

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

27. anonymous

So what are the two values we get for u?

28. anonymous

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

29. anonymous

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

30. anonymous

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

31. anonymous

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

32. anonymous

oh. limited tries?

33. 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

34. anonymous

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

35. anonymous

yep, thats what it said. thanks!

36. anonymous

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

37. anonymous

THanks for all your help! night