math question

[schneiderguy]

I'm taking a pre-calculus class in college (I took AP calc in high school but the college won't let me take the college calc class for some reason). We had a test today and the extra credit question was:

f(x) = x^4 - 4x^2 + k

For what value of k does the function have two real zeroes, each with a multiplicity of two?



So from there I figured out that since the multiplicity was 2, the function would just touch the x-axis at the zeroes, but not cross. So I took the derivative of the function and set y to 0 to find the extrema, which were +- sqrt(2). Then I plugged them back into the original equation, and set y to 0 to get k. I believe this is the correct solution, but how would you do this without derivatives? Basically, how do you get the extrema of an equation that has a degree greater than 2 without finding where the derivative of the function = 0?

 

[darkxshade]

one way would be to ask atot but you know that already

 

[Turin39789]

use the quadrature of the lune

 

[schneiderguy]

Originally posted by: darkxshade
one way would be to ask atot but you know that already

I asked google too, but it didn't want to tell me

 

[OCGuy]

Just write in .99999999.... != 1

 

[olds]

Originally posted by: Turin39789
use the quadrature of the lune

I'd have to use The Force.

 

[her209]

What's a real zero?

 

[schneiderguy]

Originally posted by: her209
What's a real zero?

a non imaginary point where the line hits the x axis.

 

[Chiropteran]

What if you just graph it?

 

[krylon]

Originally posted by: her209
What's a real zero?

A 2 dimensional sword

 

[bobsmith1492]

Graph it in Excel and look where the lines cross

Print the graph and draw lines to find the point then plug back in and tweak until the number line up

Do you need an algebraic solution??

 

[OCGuy]

Originally posted by: krylon

Originally posted by: her209
What's a real zero?

A 2 dimensional sword

:laugh:

 

[paulney]

Without using derivatives - I think you can only graph it.

 

[DrPizza]

Ummmm, factor it maybe?

(Really, it's not THAT hard!)

 

[TruePaige]

Math.

 

[DrPizza]

edit: I should probably be a little more specific...

(x-a)(x-a)(x-b)(x-b)=0

(x²-2ax+a²)((x²-2bx+b²)=0
etc.
Multiply it together, and you know there's no x^3 term, there's no x^1 term; just a constant.

 

[Hacp]

Originally posted by: DrPizza
Ummmm, factor it maybe?

(Really, it's not THAT hard!)

Teach us O wise one.

 

[DrPizza]

(x²-2ax+a²)((x²-2bx+b²)=0

x^4 -2ax^3 -2bx^3 ...
You know that -2a -2b = 0, so a = -b (there's no x^3 term)
etc.

And, the last term, a²b² has to equal k

Should be trivial to figure out from here, so I'll leave that as an exercise for the reader.

 

[eLiu]

It's just a quadratic in x^2:
let y = x^2
then your fcn = y^2 - 4y + k

Now x = +/-sqrt(y), so y^2-4y+k needs to have one positive root of multiplicity 2. Complete the square for the obvious choice, k = 4, so you have (y-2)*(y-2) = y^2 - 4y + 4 and the roots are x={-sqrt(2),-sqrt(2),sqrt(2),sqrt(2)}

edit: DrPizza's way is a lot more general so you should understand what he did. I'm just relying on 2 obvious things: 1) the expression is quadratic, 2) if the quadratic doesn't have 1 root you'll get 4 different roots, but if those things aren't obvious in the problem you're solving, then my "shortcut" is not helpful.

Edit2: OP, how would you do this WITH calculus?? I can't think of an efficient way to do it.

 

[ChAoTiCpInOy]

I tried putting it into Wolfram to no avail.

 

[OCGuy]

Its Friday evening...and you guys are discussing calculus. There are going to be some dry penises tonight.

 

[TruePaige]

Originally posted by: OCguy
Its Friday evening...and you guys are discussing calculus. There are going to be some dry penises tonight.

Probably the one attached to your asshole of a self.

 

[eLiu]

Originally posted by: OCguy
Its Friday evening...and you guys are discussing calculus. There are going to be some dry penises tonight.

It's only 3pm on the west coast, and I'm bored at work. Dry my ass.

 

[TruePaige]

Originally posted by: eLiu

Originally posted by: OCguy
Its Friday evening...and you guys are discussing calculus. There are going to be some dry penises tonight.

It's only 3pm on the west coast, and I'm bored at work. Dry my ass.

Somebody pissed in OCguy's cheerios this morning, he's been shitting in every thread he can get his hands on.

 

[frostedflakes]

Yup, just factor it. I'd also be interested to know how you'd solve this using calculus, lol. It's an algebra problem.