Figure this one out...

[theavengers85]

Dang, I forget how to do this.

If I remember correctly, quartic functions usually can't be easily factored.

[kjeopardy]

Dang, I forget how to do this.

If I remember correctly, quartic functions usually can't be easily factored.

They usually can't, but some can: x^4 - 3x^2 + 2

This factors as (x^2-2)(x^2-1) = (x^2-2)(x+1)(x-1)

[Darth Crater]

In general, you just substitute something for x^2 and see if a sensible quadratic function falls out. For example, in pi's latest post, you get u^2 - 3u + 2, which is (u + 2) (u - 1). You have to check your answers, though, because x is only one of the two square roots of u.

[kjeopardy]

In general, you just substitute something for x^2 and see if a sensible quadratic function falls out. For example, in pi's latest post, you get u^2 - 3u + 2, which is (u + 2) (u - 1). You have to check your answers, though, because x is only one of the two square roots of u.

Exactly: you can solve higher degree equations through substituting and making them into a quadratic. We can generalize this to equations other than the quartic. Try equations of the form x^2n + x^n + c, for some constant c.

[kjeopardy]

I have a new problem...this problem is deceptively easy, yet ridiculously frustrating...I'm not sure if my solution is correct, so I'll hope for some feedback. Note that I didn't invent this. Here we go:
______________________________________________
Given that:

a^2 + b^2 = 1

c^2 + d^2 = 1

ac + bd = 0

Show that:

a^2 + c^2 = 1

b^2 + d^2 = 1

ab + cd = 0

[Nathan_Mates]

a = 1/2^(1/2), b = 1/2^(1/2), c = 1/2^(1/2), and d = -1/2^(1/2).

[kjeopardy]

a = 1/2^(1/2), b = 1/2^(1/2), c = 1/2^(1/2), and d = -1/2^(1/2).

Interesting...did you solve the system? I'm not sure this proves anything. I see your point though: (1/2)^(1/2) squared times 2 is 1...

[Nathan_Mates]

Yeah, I know it's not exactly what you are asking for. I just didn't feel like going through all the algebra to solve the proof, so I just mentally solved for the values that would make all of the equations listed true for fun and to give my mind some excercise.

[kjeopardy]

Yeah, I know it's not exactly what you are asking for. I just didn't feel like going through all the algebra to solve the proof, so I just mentally solved for the values that would make all of the equations listed true for fun and to give my mind some excercise.

Understood. However, a very simple and elegant proof exists. If you do it right, there is no need for lots of algebra. This is part of the challenge I guess...

[11_Panama_]

Did you guys get beat up alot?