# Polynomial find the range of root

1. Aug 14, 2004

### scoutfai

The question is : Show that one of the root of the equation y^3 - 3y + 4 = 0 lies between -3 and -2.

REMEMBER : If you want to use some statement, Law, identity or logic, please PROVE it first before you proceed to answer the question..

Difficulty for me is, normally we are asked to solve for the exact value of the roots when given an equation, but now we are asked to find the range for one of the roots, hence i feel so weird and have no idea how to continute.
Any expert, if you know, please teach me, i appreciate u help, thanks you.

2. Aug 14, 2004

### Crumbles

Ok, imagine that a function f(y)=y^3 - 3y + 4

The question is asking at what point that function equals zero. (which is what "the root of the equation" means. The easiest way to understand how to solve this problem is to plot the function f(y) against y on graph for some values between -3 and 2. You will find that somewhere in this range your graph will cut the x-axis. The x-axis value at this point is known as the root of the equation.

Now to prove this mathematically, all you need to do is find the value of f(y) for y=-3 and then find the value of f(y) for y=2. You will find that the answers to f(-3) and f(2) have opposite signs, which graphically means that the point of your curve for f(-3) lies on the opposite side of the x-axis than the point of your curve for f(2).

Now if you think about it, if the curve is above the x axis at one point and below the x axis at the other, it must mean that somewhere between those two values (-3 and 2) your curve cuts the x axis!

Remember that where the curve cuts the x axis is where the root is (this is the point where f(y) = 0). Therefore, the root must be between -3 and 2.

3. Aug 14, 2004

### humanino

Let $$f(y) = y^3 -3y+4$$
f is defined over the whole real axis, and is continuous everywhere because it is a polynomial. Now $$f(-3)<0$$ and $$f(-2)>0$$. The function has to vanish somewhere in between.

Think about the curve representing the function. It has to cross the axis somewhere.

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Crumbles was faster than me, and brought a better explanation... :uhh:

Last edited: Aug 14, 2004