Question

find all the zeros of the equation -3x^4+27x^2+1200=0

Answer 1

Let \(\large\color{black}{ x^2=a }\)

Answer 2

I am going to give you an example, please don't leave, okay?

Answer 3

okay so x^4= x^2?

Answer 4

you mean that \(\large\color{black}{ x^4=a^2}\) ?

Answer 5

First divide all terms by -3

Answer 6

yeah i guess haha

Answer 7

This will allow for an easier factoring of the quartic function.

Answer 8

So, then your equation becomes:
\(\large\color{black}{ -3a^2+27a+1200=0}\)

Answer 9

then as Jhanny adivsed, take out -3.

Answer 10

-1a^2+9a+400=0?

Answer 11

\[-3x^4+27x^2+1200=0\]\[\to x^4 -9x^2 -400=0\]Find two factors that multiply to give you 400 and add to give you -9

Hint: think of quarters.

Answer 12

oh divide by -3!

Answer 13

yes, @jasmine61 by -3, and what do you get then?

Answer 14

x^4-9x^2-400=0??

Answer 15

you could also use the method @SolomonZelman mentioned above, and let \(u=x^2\)

Answer 16

I said, a, not u :P

Answer 17

Yes you are right, @jasmine61

@SolomonZelman , same thing.

Answer 18

okay good! im getting the hang of it

Answer 19

You will have \[u^2 -9u-400=0\]

Answer 20

It's not. because "with u" I think of integrals, with "a" I think of algebra, because that is how my teacher taught me a long time ago. with a, not u.

Answer 21

u and a and x and y are all just numbers with letter representatives. I guess it is how you look at it and define how you solve things.

Answer 22

well, I already started from making it a, why do we need to change it?

Answer 23

Hell, you can even do a p substitution :P

Answer 24

whats the next step? hahah

Answer 25

Ok, let's use `a` then

Answer 26

\(\large\color{black}{ a^2-9a-400=0}\)

Answer 27

the next step is to find two numbers that multiply to give you -400 and add to give you -9

Answer 28

And then you know that:
\(\large\color{black}{ 400\div 16=25}\) and \(\large\color{black}{ (-25)+16=-9}\)

Answer 29

sorry gave it away.

Answer 30

Yeah, I said think of quarters for a reason :P

Answer 31

so is it 25 and -16 or am i totally off?

Answer 32

Well, if you have positive 25 and negative 16, then your middle term, `-9` would become `+9` because you're letting 25 be positive, and 25-16 = 9, not -9

Answer 33

the other way

Answer 34

oh okay, but i thought you have to set them to 0?

Answer 35

You will, once you write it in factored form.

Answer 36

oh oh okay, i understand now!

Answer 37

You have -25 and +9 as your factors, how will you write \(a^2 -9a-400=0\) in factored form?

Answer 38

(a-25)(a+9)??

Answer 39

Good :) that is right. \[(a-25)(a+9)=0\]Now we resubstitute our \(a\) value.

Answer 40

What did @SolomonZelman say \(a\) was equal to?

Answer 41

x^2

Answer 42

Good, so we will now have : \((x^2-25)(x^2+9)=0\)

Answer 43

We factor each of these terms separately.\[x^2-25 = 0\]\[x^2+9=0\]

Answer 44

Are you able to solve for both of these?