# Approximate all real roots of the equation to two decimal places using Newton's Method. X^4 = 125Must use Newtons Method. Thanks!

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Approximate the real roots of `x^4=125` to 2 decimal places:

Consider the graph of `x^4-125=0` :

It appears the roots are a littleabove 3 and below -3.

To use Newton's method,we find a"good" initial approximation, `x_1` . Then we determine a new approximation `x_(n+1)=x_n-(f(x_n))/(f'(x_n))` . We continue this process until `|x_n-x_(n+1)|` is within the desired accuracy; in this case to two decimal places.

(1) The root near 3: Let `x_1=3.5` . Note that `f'(x)=4x^3` . Then:

`x_2=3.5-(3.5^4-125)/(4(3.5^3))~~3.35386`

`x_3=3.35386-(3.35386^4-125)/(4(3.35386^3))~~3.34374`

`x_4=3.34374-(3.34374^4-125)/(4(3.34374^3))~~3.34370`

**Since `|3.34374-3.34370|=0.00004` is within the desired accuracy we have thefirst root is approximately 3.34**

(2) The root near -3. Let `x_1=-3.5`

`x_2=-3.5-((-3.5)^4-125)/(4((-3.5)^3))~~-3.35386`

`x_3=-3.35386-((-3.35386)^4)/(4((-3.35386)^3))~~-3.34374`

`x_4=-3.34374-((-3.34374)^4-125)/(4((-3.34374)^3))~~-3.34370`

**So the second root is approximately -3.34**

**Sources:**