Babynini
  • Babynini
Prove the equation has at least 1 real root.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Babynini
  • Babynini
#56 :)
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ganeshie8
  • ganeshie8
Have you tried using IVT ?
Babynini
  • Babynini
Noo. I'm not sure where to start.

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Babynini
  • Babynini
I don't know what IVT is o.o
dan815
  • dan815
find that the function has a positive and a negative value
dan815
  • 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
  • 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
  • dan815
|dw:1444528198685:dw|
Babynini
  • Babynini
Ahh k. ^ makes sense thus far.
ganeshie8
  • 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
  • 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
  • Babynini
degree of 5? hm?
ganeshie8
  • ganeshie8
Look at the equayion in problem #56 first convinvce yourself that it is a polynomial equation
dan815
  • dan815
just do this f(10000) = positive f(-1000000) = negative you are done
dan815
  • 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
  • dan815
the curve* must pass if u are picky -.-
Babynini
  • Babynini
Haha k. Well, what does it mean for it to have a real root? that just means it passes through 0?
dan815
  • dan815
thats what a root is
dan815
  • dan815
a real root means u have an x value where that intersects the x axis
ganeshie8
  • ganeshie8
real roots are same as x intercepts in the graph
Babynini
  • 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
  • ganeshie8
for example, below graph has 3 x intercepts, so we say the number of real roots are 3 |dw:1444528871031:dw|
dan815
  • dan815
highest is odd degree* polynomial
dan815
  • dan815
even highest degree polynomials dont need to have real root
Babynini
  • Babynini
I see, I see. Ok, this is making much more sense.
Babynini
  • 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
  • dan815
you might need more support, the better way is to show that htere is a positive and a negative value
dan815
  • dan815
and say since its continuous.. it must pass thru xaxis
dan815
  • 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
  • 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
  • dan815
pale*
Babynini
  • Babynini
hm ok. Should I do any equations to prove it?
dan815
  • 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
  • dan815
so u have shown that this functions goes to inf and -inf at its end points
Babynini
  • Babynini
hmm ok. Why did you add all those extra numbers? o.o
dan815
  • dan815
the abcd?
Babynini
  • Babynini
Yeah. If the original is: x^5-x^2+2x+3
dan815
  • dan815
its just more general, u can plug in the values for a b c d respectively
dan815
  • dan815
i couldnt remember your question exactly so
Babynini
  • Babynini
^ thanks so much. I think I get it :) looking over notes and looking at all you said.

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