Can someone help me witht he root in the interval: f(x)=x^4+3x^3-5x^2-6 on (1,2)

Coachdwhite02 said:
f(x)= x^{4} +3x^{3}-5x^{2}-6 has a root in the interval (1,2).
Click to expand...
First, by the "rational root theorem", only 1, -1, 2, -2, 3, -3, 6, and -6 can be rational number zeros for this expression (technically, an equation has "roots". An expression has "zeros" that are the roots of that expression set equal to 0.)
but evaluating at each of those shows none is a zero. So any such zero is an irrational number.

There is a "quartic formula" (http://www.sosmath.com/algebra/factor/fac12/fac12.html) but that is difficult to use. You might be better to use a numerical method. The simplest, though also tedious, is "bisection". You know that there is a root between 1 and 2, because f(1)= 1+ 3- 5- 6= -7, negative, while f(2)= 16+ 24- 20- 6= 14, positive, but you do not know where. So try half way between: x= 3/2. f(3/2)= -2.0625, negative. So there must be a zero between 3/2 and 2. Try half way between, x= 7/4. f(7/4)= 4.14453125, positive. So there must be a zero between 3/2 and 7/4. Try half way between, x= 13/8. f(13/8)= 0.6428. That's positive (but closer to 0) so there must be a zero between 3/2 and 13/8. Keep doing that until you get the value sufficiently close to 0.
 
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Coachdwhite02 said:
f(x)= x^{4} +3x^{3}-5x^{2}-6 has a root in the interval (1,2).
Click to expand...
What is the question?

Is it to prove that there is a root in the interval or is it to approximate the root by Newton's method or to find the exact root or what?

In the future, please give the problem exactly and completely.
 
Coachdwhite02 said:
f(x)= x^{4} +3x^{3}-5x^{2}-6 has a root in the interval (1,2).
Click to expand...

The graph of \(\displaystyle f(x)= x^{4} +3x^{3}-5x^{2}-6\) is continuous over the interval.

f(1) = 1+3-5-6=-7 < 0

f(2) = 24 + 3*23 - 5*22 -6 = ????

What can you conclude if f(1) and f(2) are different signs?? (Think graphically.)

Finding the root between 1 and 2 is a much harder problem.
 
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