Question

Can someone please walk me through and/or guide me on how to find all the zeroes of this polynomial equation? I understand the basics of finding rational roots and synthetic division but am confused on how to approach this with such a large constant. -3x^4+27X^2+1200=0

Answer 1

@jim_thompson5910 any chance you could walk me through this?

Answer 2

Notice how we have a 4th power as the largest exponent, it skips the 3rd power term, goes to an x^2 term and skips the next x term

Basically all the exponents are even here

Answer 3

Yes, and since it is to the 4th power I know there are 4 roots.

Answer 4

since we have relatively nice format, we can use this trick

let z = x^2

squaring both sides gives z^2 = (x^2)^2 = x^(2*2) = x^4

so if z = x^2 then z^2 = x^4

agreed so far?

Answer 5

I see what you are getting at but am a little confused on this trick you are using.

Answer 6

Why on earth do this? Well because we can transform -3x^4+27X^2+1200=0 into something a bit more familiar


Using
z = x^2
z^2 = x^4
we can make replacements


-3x^4+27X^2+1200=0
will turn into
-3z^2+27z+1200=0

Answer 7

Oh so all you are doing is squaring those 2 terms. I see now

Answer 8

at this point, you probably know how to solve -3z^2+27z+1200=0 right?

Answer 9

My problem is with the large constant. I'm unsure how to approach it. I know I could find rational roots but there are so many when dealing with 1200, or I could try to find factors of 1200 that add to 27.

Answer 10

why not use the quadratic formula?

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 11

sure you could guess and check, but you may be there a while

Answer 12

Oh I see! I forgot about that.

Answer 13

I should be able to find the roots based on that. Haven't used that formula in a but but think it is straightforward..

Answer 14

once you figure out the two solutions for z, use them to find what x must equal

Answer 15

Can I not simply plug in -3z^2+27z+1200=0 to the formula? The whole z coefficient is a brand new concept for me.

Answer 16

think of it as -3x^2+27x+1200=0

Answer 17

in your case, a = -3, b = 27, c = 1200

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-27\pm\sqrt{27^2-4*(-3)*1200}}{2*(-3)}\]
I'll let you keep going

Answer 18

So I simplified and got x = -16 and x=25

Answer 19

those are the z values

z = -16 and z = 25

Answer 20

if z = -16,then what is x?

recall that we let z = x^2

Answer 21

so x = 256 and x = 625?

Answer 22

z = -16
z = x^2
-16 = x^2
x = ?? or x = ??

Answer 23

Oh silly me I did the opposite and raised it to the power of 2 my bad. Instead I needed to take the square root which is 4i and 25 is 5

Answer 24

don't forget the plus minus

Answer 25

x^2 = -16
x = 4i or x = -4i

Answer 26

Yep, and x =5 or x= -5

Answer 27

good

Answer 28

so as expected, we end up with 4 complex solutions

Answer 29

Yes which makes sense since the first term had a degree of 4

Answer 30

Thank you so much for your time and help on this it is greatly appreciated! People like you make Open Study a great place to ask questions and encourage me to get help.

Answer 31

yes
largest exponent = 4
degree = 4
number of solutions (real or imaginary) = 4

Answer 32

I'm glad I could help out

Answer 33

I'm going to write you a testimonial, you are great! I am impressed at how fast you answered this and the methods you used to help guide me to the answer.

Answer 34

Thank you