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An approximate solution to an equation is found using this iterative process.
(x₂)³-5
and x₁ = -1
10
Xn+1 =
-0.6
a) (i) Work out the value of x₂
(ii) Work out the value of X3
b) Work out the solution to 6 decimal places.
-0.5216

1 Answer

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Final answer:

The iterative process is used to approximate the solution to an equation. The values of x₂ and x₃ are found using the given formula. To find the solution to 6 decimal places, we continue the iterative process.

Step-by-step explanation:

In this problem, we are given an iterative process to approximate the solution to an equation. We start with x₁ = -1 and use the given formula to find x₂ and x₃.

(i) To find x₂, we substitute x₁ = -1 into the formula: x₂ = (-1)³ - 5/10 = -1 - 0.5 = -1.5.

(ii) To find x₃, we substitute x₂ = -1.5 into the formula: x₃ = (-1.5)³ - 5/10 = -3.375.

(b) To find the solution to 6 decimal places, we can continue the iterative process using the given formula until we reach the desired accuracy. Based on the given answer, the approximate solution to 6 decimal places is -0.5216.

User Annie Sheikh
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