Question

Prove the equation has at least 1 real root.

Answer 1

#56 :)

Answer 2

Have you tried using IVT ?

Answer 3

Noo. I'm not sure where to start.

Answer 4

I don't know what IVT is o.o

Answer 5

find that the function has a positive and a negative value

Answer 6

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

Answer 7

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

Answer 8

Ahh k. ^ makes sense thus far.

Answer 9

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.

Answer 10

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

Answer 11

degree of 5? hm?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

thats what a root is

Answer 18

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

Answer 19

real roots are same as x intercepts in the graph

Answer 20

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?

Answer 21

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