Question

@mathmale Hello, could you help me with finding all the zeros of the function?

y= (x+3)(x+2)(x-2)

(Is it -3, -2, 2?)

Answer 1

Hi, N,
I'd be happy to help. At the moment I'm working with another student. I'll get back to you asap, but encourage you to "hear" what others on Open Study have to contribute also.

Answer 2

@Saphira_Shapiro Huh?

@mathmale Okay, thank you!

Answer 3

@Nicole143 you done it right,

for the left hand side to be zero,

at least one of the factors on the right hand side must be zero

so you did
(x+3)=0
or
(x+2)=0
or
(x-2)=0

which you simplified to get x= -3, -2, or 2 √√√

Answer 4

Awesome, could you help me with another? And would you like me to open it in another question?

Answer 5

sure,

Answer 6

@UnkleRhaukus Okay, so I have to find all the zeros of the equation 27x^2 - 324 = -x^4

Answer 7

first get all the terms on the right hand side,


then i would substitute x^2=z, x^4=z^2

to get a quadratic

Answer 8

Okay, I now have 27z + z^2 - 324 = 0

Answer 9

can you factor that , with the quadratic equation?

Answer 10

With the equation x = -b+- the square root of b^2 - 4ac / 2a?

Answer 11

yeah that's the quadratic formula, can you get the factors now?

Answer 12

I don't know.. I don't think it's quite clicking.

Answer 13

actually you dont need to explicitly find the factors, the quadratic formula will give you the zeros straight away

Answer 14

So just put in 27z for a, z^2 for b, and -324 for c?

Answer 15

z^2+27z - 324 = 0

a=1 b=27 c=-324

Answer 16

Oh, okay. One moment.

Answer 17

I now have

Answer 18

I didn't hit post:/
I have x = -27 +- the square root of 2592 / 2

Answer 19

Or are you supposed to leave for as a positive to be able to minus?

Answer 20

z=[-b±√(b^2-4ac)]/2a

=[-27±√(27^2-4(1)(-324))]/2
=[-27±√(2025)]/2

Answer 21

Are you still working on the original problem ("Hello, could you help me with finding all the zeros of the function?

y= (x+3)(x+2)(x-2) )?

Need me, or are you two getting where you want to go?

Answer 22

we done that one already @mathmale

Answer 23

I would love for you to help as well - I'm a bit off on my answer at least for what he has. And yes we are on a different one now.

Answer 24

Great!

Nicole, I need to get off the computer very soon. But you have my address in case you want to share future problems with me.

Answer 25

I unfortunately need to get this done tonight but hopefully someone will stick around and help(: @mathmale

Answer 26

I see that U-R has explained some things quite clearly and competently. What specific part of your discussion is not yet clear for you?

Answer 27

Will you be around long enough to explain a different question to me? @mathmale
For this one I just didn't get the sam answer as him..

Answer 28

*same

Answer 29

i kinda have to go now, so you can help with this second one @mathmale

dont forget when you get the z's you still need to get the x's by taking ±sqrts

Answer 30

Nicole, I am preparing to leave tomorrow morning on a 1,500 mile trip to Canada. However, if you'll post the next question I'll get back to you after a few minutes away from my computer.

Thank you very much U-R!

Answer 31

Okay, thank you for your help! @UnkleRhaukus

Answer 32

@mathmale Okay(:

Answer 33

Nicole, either post your question here, or email it to me. Your choice. Or, send it me as a message.

Answer 34

I'll email you.

Answer 35

@mathmale I'll send it in a few minuets, thank you for the help!