Skip to content# Iteration Equations: Comprehensive Guide with Worksheet

## What is Iteration?

## Steps for Solving an equation using iteration

## Steps to locate the roots of an equation between two points

## Practice Questions

### Solutions:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

**Iteration**

- Iteration is a mathematical process of repeatedly refining a solution or estimate to a problem.
- It is a valuable tool for finding numerical solutions to equations that cannot be solved analytically.

In this article, we will discuss:

**What is Iteration?****Steps for Solving an equation using iteration****Steps to locate the roots of an equation between two points**

Here is one more link to practice a few extra questions: Maths Genie Iteration Questions

- Iteration is a mathematical process that involves repeating a specific sequence of steps until a satisfactory solution or estimate is achieved.
- The main idea behind iteration is to use the previous solution to generate a new one, which should be closer to the actual answer.
- This process is repeated until the desired level of accuracy is achieved.
**Iteration is commonly used to solve nonlinear equations, such as transcendental or polynomial equations, that cannot be solved analytically.**- It can also be used to find the roots of a function or estimate the value of an integral.

- The following steps can be used to solve an equation using iteration:

**Step #1:** Rearrange the equation to get it into the form **f(x) = 0,** where **f(x)** is a continuous function.

**Step #2:** Select an initial guess, **x₀**, for the solution.

**Step #3:** Apply the iteration formula **x _{n+1} = g(x_{n}),** where g(x) is a function that generates a new solution using the previous solution.

**Step #4:** Repeat step #3 until the desired level of accuracy is achieved.

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**Solved Example: **

**Question 1: Solve the equation 3x – x ^{3} = -11 using iteration.**

Solution:

**Step #1:**Rearrange the equation to get it into the form f(x) = 0. We get

**f(x) = 3x – x ^{3} + 11.**

**Step #2:**Select an initial guess, x₀, for the solution. Let**x₀ = 3.**

**Step #3:**Apply the iteration formula:

which generates a new solution using the previous solution.

**Step #4:**Repeat step #3 until the desired level of accuracy is achieved.

**x _{1} = (11 + 3x_{0})^{1/3} = (11 + 3x_{32})^{1/3} = 2.7144**

**x _{2} = (11 + 3x_{1})^{1/3} = (11 + 3x_{2.7144})^{1/3} = 2.6751**

**x _{3} = (11 + 3x_{2})^{1/3} = (11 + 3x_{2.6751})^{1/3} = 2.6696**

**x _{4} = (11 + 3x_{3})^{1/3} = (11 + 3x_{2.6696})^{1/3} = 2.6688**

**x _{5} = (11 + 3x_{4})^{1/3} = (11 + 3x_{2.6688})^{1/3} = 2.6687**

After several iterations, we get x_{5} ≈ 2.6687

**Question 2: Solve the equation e ^{x} – 4x = 0 using iteration.**

Solution:

**Step #1:**Rearrange the equation to get it into the form f(x) = 0. We get

**f(x) = e ^{x} – 4x.**

**Step #2:**Select an initial guess, x₀, for the solution. Let**x₀ = 1.**

**Step #3:**Apply the iteration formula

**xᵢ = ln(4xᵢ),**

which generates a new solution using the previous solution.

**Step #4:**Repeat step #3 until the desired level of accuracy is achieved.

**Iteration 1: x₁ = ln(4 x 1) = ln(4) = 1.386294361119891**

**Iteration 2: x₂ = ln(4 x 1 _{3863}) = 1.712932688682084**

**Iteration 3: x₃ = ln(4 x 1 _{7129}) = 1.924482201641693**

**Iteration 4: x₄ = ln(4 x 1 _{9245}) = 2.040960554861554**

**Iteration 5: x _{5} = ln(4 x 2_{0410}) = 2.099734244947598**

**Iteration 6: x _{6} = ln(4 x 2_{0997}) = 2.128088838501357**

At this point, we have achieved the desired level of accuracy and can stop iterating.

After several iterations, we get x_{6} ≈ 2.128

- The following steps can be used to locate the roots of an equation between two points:

**Step #1:** Find the value of the function at the two intervals a and b.

**Step #2:** Check if there is a sign change between the two values obtained in step #2.

If the signs are different, there is a root between the two intervals.

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**Solved Example: **

**Question: ****Find the root of the equation f(x) = x ^{3} – 4x^{2} + 5x – 2 between the**

Solution:

**Step #1:**Find the value of the function at the two intervals a = 0 and b = 2.**f(0) = -2 and f(2) = 2****Step #2:**Check if there is a sign change between the two values obtained in step #2.

**Since there is a sign change between -2 and 2, there is a root between the intervals [0, 2].**

**Question 1: **Solve the equation x^{3} – 2x – 5 = 0. Take the initial value as x = 2.

**Question 2: **Solve the equation 2x^{2} – 5x + 3 = 0. Take the initial value as x = 1.

**Question 3: **Solve the equation e^{x} – x – 2 = 0. Take the initial value as x = 1.

**Question 4: **Solve the equation 4cos(x) – x = 0. Take the initial value as x = 1.

**Question 5: **Solve the equation ln(x) – x + 2 = 0. Take the initial value as x = 2.

**Question 6: **Solve the equation x^{2} – x – 1 = 0. Take the initial value as x = 2.

**Question 7: **Solve the equation 3x – 2^{x} = 0. Take the initial value as x = 1.

**Question 8: **Solve the equation sin(x) + x – 2 = 0. Take the initial value as x = 1.

**Question 9: **Solve the equation x^{3} + x + 1 = 0. Take the initial value as x = 0.

**Question 10: **Solve the equation 5x^{2} – 6x – 2 = 0. Take the initial value as x = 1.

**Question 1:** Solve the equation x^{3} – 2x – 5 = 0. Take the initial value as x = 2.

Solution:

Using the method of iteration, we can start with an initial value of x = 2 and iterate using the formula:

**xₙ₊₁ = ∛(xₙ² + 5/2)**

**Iteration 1:**

**x₁ = ∛(2² + 5/2) ≈ ∛(4 + 5/2) ≈ ∛(8.5) ≈ 2.0801**

**Iteration 2:**

**x₂ = ∛(2.0801² + 5/2) ≈ ∛(4.3264 + 5/2) ≈ ∛(7.3264) ≈ 2.1259**

Continuing the iterations, we can approach the solution of the equation.

**Question 2:** Solve the equation 2x^{2} – 5x + 3 = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

**xₙ₊₁ = (5xₙ – 3)/(2xₙ)**

**Iteration 1:**

**x₁ = (5(1) – 3)/(2(1)) = (5 – 3)/2 = 2/2 = 1**

**Iteration 2:**

**x₂ = (5(1) – 3)/(2(1)) = (5 – 3)/2 = 2/2 = 1**

Continuing the iterations, we find that x remains 1, indicating that it is the solution to the equation.

**Question 3:** Solve the equation e^{x} – x – 2 = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

**xₙ₊₁ = ln(xₙ + 2)**

**Iteration 1:**

**x₁ = ln(1 + 2) = ln(3) ≈ 1.0986**

**Iteration 2:**

**x₂ = ln(1.0986 + 2) ≈ ln(3.0986) ≈ 1.1314**

Continuing the iterations, we can approach the solution of the equation.

**Question 4:** Solve the equation 4cos(x) – x = 0. Take the initial value as x = 1.

Solution:

Using the method of iteration, we can start with an initial value of x = 1 and iterate using the formula:

**xₙ₊₁ = arccos(xₙ/4)**

**1:**

**x₁ = arccos(1/4) ≈ 1.3181**

**Iteration 2:**

**x₂ = arccos(1.3181/4) ≈ 1.2863**

Continuing the iterations, we can approach the solution of the equation.

**Question 5:** Solve the equation ln(x) – x + 2 = 0. Take the initial value as x = 2.

Solution:

Using the method, we can start with an initial value of x = 2 and iterate using the formula:

**xₙ₊₁ = e ^{(x}^{ₙ – 2)}**

**Iteration 1:**

**x₁ = e ^{(2 – 2) }= e^{0} = 1**

**Iteration 2:**

**x₂ = e ^{(1 – 2)} = e^{-1} ≈ 0.3679**

Continuing the iterations, we can approach the solution of the equation.

**Question 6:** Solve the equation x^{2} – x – 1 = 0. Take the initial value as x = 2.

Solution:

Using the method, we can start with an initial value of x = 2 and iterate using the formula:

**xₙ₊₁ = 1 + 1/xₙ**

**Iteration 1:**

**x₁ = 1 + 1/2 = 1.5**

**Iteration 2:**

**x₂ = 1 + 1/1.5 = 1.6667**

Continuing the iterations, we can approach the solution of the equation.

**Question 7:** Solve the equation 3x – 2^{x} = 0. Take the initial value as x = 1.

Solution:

Using the method, we can start with an initial value of x = 1 and iterate using the formula:

**xₙ₊₁ = log₂(3xₙ)**

**Iteration 1:**

**x₁ = log₂(3(1)) = log₂(3) ≈ 1.585**

**Iteration 2:**

**x₂ = log₂(3(1.585)) ≈ log₂(4.755) ≈ 2.246**

Continuing the iterations, we can approach the solution of the equation.

**Question 8:** Solve the equation sin(x) + x – 2 = 0. Take the initial value as x = 1.

Solution:

**xₙ₊₁ = 2 – sin(xₙ)**

**Iteration 1:**

**x₁ = 2 – sin(1) ≈ 1.158**

**Iteration 2:**

**x₂ = 2 – sin(1.158) ≈ 1.603**

Continuing the iterations, we can approach the solution of the equation.

**Question 9:** Solve the equation x^{3} + x + 1 = 0. Take the initial value as x = 0.

Solution:

Using the method of iteration, we can start with an initial value of x = 0 and iterate using the formula:

**xₙ₊₁ = -1/(xₙ ^{2} + 1)**

**Iteration 1:**

**x₁ = -1/(0 ^{2} + 1) = -1/1 = -1**

**Iteration 2:**

**x₂ = -1/((-1) ^{2} + 1) = -1/2**

Continuing the iterations, we can approach the solution of the equation.

**Question 10:** Solve the equation 5x^{2} – 6x – 2 = 0. Take the initial value as x = 1.

Solution:

**xₙ₊₁ = (5xₙ ^{2} – 2) / 6**

**Iteration 1:**

**x₁ = (5(1) ^{2} – 2) / 6 = (5 – 2) / 6 = 3/6 = 0.5**

**Iteration 2:**

**x₂ = (5(0.5) ^{2} – 2) / 6 ≈ (5(0.25) – 2) / 6 ≈ (1.25 – 2) / 6 ≈ -0.125**

Continuing the iterations, we can approach the solution of the equation.

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