# How do you use Newton's Method to approximate the root of the equation x^4-2x^3+5x^2-6=0 on the interval [1,2] ?

Sep 21, 2014

The answer is $1.217562155$.

Recall that Newton's Method uses the formula:

x_(n+1)=x_n−(f(x_n))/(f'(x_n))

The equation is already a function, so:

$f \left(x\right) = {x}^{4} - 2 {x}^{3} + 5 {x}^{2} - 6$

And we need the derivative:

$f ' \left(x\right) = 4 {x}^{3} - 6 {x}^{2} + 10 x$

The easiest way to iterate is to program your calculator. Enter $f \left(x\right)$ into ${Y}_{1}$ and $f ' \left(x\right)$ into ${Y}_{2}$. Then enter a very short program that does this:

A−(Y_1(A))/(Y_2(A))->A

You can go to my website for specific instructions for the TI-83 or the Casio fx-9750 .

Finally, you need a starting value, ${x}_{1}$. Since the question is asking for a root in the interval $\left[1 , 2\right]$, we should start at $x = 1.5$. So enter the following into your calculator,

$1 \to A$

Then execute the program until you get the desired accuracy:

$1.2625$
$1.218807774$
$1.217563128$
$1.217562155$
$1.217562155$

We get 3 digits of accuracy after 2 iterations, 6 after 3 iterations, and 10 after 4 iterations. So the answer converges very quickly for this root.