Prove the equation has at least 1 real root.

- Babynini

Prove the equation has at least 1 real root.

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- schrodinger

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- Babynini

#56 :)

##### 1 Attachment

- ganeshie8

Have you tried using IVT ?

- Babynini

Noo. I'm not sure where to start.

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## More answers

- Babynini

I don't know what IVT is o.o

- dan815

find that the function has a positive and a negative value

- dan815

or use the fact that it's an odd functions, so the ends, have one going to infnite and the other going to negative infinity

- dan815

and thsi is an analytical function meaning there are no holes, and its connected so it must have passed through 0 to do this

- dan815

|dw:1444528198685:dw|

- Babynini

Ahh k. ^ makes sense thus far.

- ganeshie8

If you're not familiar with IVT yet, you may also simply use the fundamental theorem of algebra :
A polynomial of degree "n" has exactly "n" roots and out of these "n" roots, the complex roots always come in conjugate pairs.

- ganeshie8

since the polynomial in question has a degree of 5, which is odd, it follows that at least one root must be real.

- Babynini

degree of 5? hm?

- ganeshie8

Look at the equayion in problem #56
first convinvce yourself that it is a polynomial equation

- dan815

just do this f(10000) = positive
f(-1000000) = negative
you are done

- dan815

since its a continuos function these 2 points are connected in between in some way, and the line must obviously pass y=0

- dan815

the curve* must pass if u are picky -.-

- Babynini

Haha k. Well, what does it mean for it to have a real root? that just means it passes through 0?

- dan815

thats what a root is

- dan815

a real root means u have an x value where that intersects the x axis

- ganeshie8

real roots are same as x intercepts in the graph

- Babynini

Ahhh k. So as soon as I recognize it's a polynomial then I basically know it has root? -.- why do I want to go about proving it then?

- ganeshie8

for example, below graph has 3 x intercepts, so we say the number of real roots are 3
|dw:1444528871031:dw|

- dan815

highest is odd degree* polynomial

- dan815

even highest degree polynomials dont need to have real root

- Babynini

I see, I see. Ok, this is making much more sense.

- Babynini

So for part a, I legit just have to say: "since the polynomial in question has a degree of 5, which is odd, it follows that at least one root must be real."

- dan815

you might need more support, the better way is to show that htere is a positive and a negative value

- dan815

and say since its continuous.. it must pass thru xaxis

- dan815

you cant just be saying oh its degree 5, so it just works, u are going to have to link polynomial functiosn to their odd function counter parts

- dan815

you are going to have to state something like
since its a composition of odd and even functions
x^5, ax^4, bx^3,cx^3,dx,3
as x goes to infinite the additions of the other degrees will pare in comparision and the end points will exhibit the odd degree characteristics

- dan815

pale*

- Babynini

hm ok. Should I do any equations to prove it?

- dan815

u could do
x^5 + ax^4 + bx^3 + cx^3 + dx +3 =0
divide everything by x^4
x+a+b/x + c/x^2 + d/x^3 + 3/x^4=0
now u can check x=inf and -inf
if u set x=inf ull end up with inf +a
if u set x=-inf ull end up with -inf +a

- dan815

so u have shown that this functions goes to inf and -inf at its end points

- Babynini

hmm ok. Why did you add all those extra numbers? o.o

- dan815

the abcd?

- Babynini

Yeah. If the original is: x^5-x^2+2x+3

- dan815

its just more general, u can plug in the values for a b c d respectively

- dan815

i couldnt remember your question exactly so

- Babynini

^ thanks so much. I think I get it :) looking over notes and looking at all you said.

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