GCSE Iteration Practice Questions

Free GCSE Iteration practice questions with full step-by-step worked solutions. Covers substitution, evaluating f(x), sign of f(x), iterative formula. Practise exam-style problems and check your method.

substitutionevaluating f(x)sign of f(x)iterative formularoundingsign change
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
f(x)=x3−2x−5f(x) = x^3 - 2x - 5. Work out the value of f(2)f(2).
Show worked solution

Worked solution

  1. Substitute the value into the function

    f(2)=23−2×2−5f(2) = 2^{3} - 2 \times 2 - 5

    Replace every xx in f(x)f(x) with 22.

  2. Work out each term

    f(2)=8−4−5f(2) = 8 - 4 - 5

    Deal with the powers first, then the multiplications, then add and subtract.

  3. State the value

    f(2)=−1f(2) = -1

    So f(2)f(2) is -1, which is negative.

Answer
f(2)=−1f(2) = -1
Question 2
2 markseasy
The equation x2−4x−3=0x^2 - 4x - 3 = 0 is to be solved by iteration. Which of these iterative formulae is a correct rearrangement of the equation?
Show worked solution

Worked solution

  1. Rearrange the equation

    x2=4x+3x^2 = 4x + 3

    Move 4x+34x + 3 to the right-hand side.

  2. Undo the power

    x=4+3xx = 4 + \frac{3}{x}

    Divide every term by xx.

  3. Write it as an iterative formula

    xn+1=4+3xnx_{n+1} = 4 + \frac{3}{x_n}

    The new value xn+1x_{n+1} comes from putting the old value xnx_n into the right-hand side.

Answer
xn+1=4+3xnx_{n+1} = 4 + \frac{3}{x_n}
Question 3
2 marksintermediate
The iterative formula xn+1=4xn+1x_{n+1} = \frac{4}{x_n + 1} converges to a number xx. This number is a root of which equation?
Show worked solution

Worked solution

  1. Write down what the limit satisfies

    x=4x+1x = \frac{4}{x + 1}

    If the iterates settle on a value xx, then feeding xx in gives xx back out.

  2. Clear the root or the fraction

    x(x+1)=4x(x + 1) = 4

    Multiply both sides by (x+1)(x + 1).

  3. Rearrange into the form f(x)=0f(x) = 0

    x2+x−4=0x^2 + x - 4 = 0

    Expand the bracket and take the 4 across.

  4. State the equation

    x2+x−4=0x^2 + x - 4 = 0

    This is the equation the iteration is solving.

  5. Check numerically

    x5=1.617021…x_5 = 1.617021\ldots

    Running the iteration from x0=1x_0 = 1 gives x5≈1.617021x_5 \approx 1.617021.

  6. Substitute the limit back in

    f(1.617021)≈0.231779f(1.617021) \approx 0.231779

    The value makes f(x)f(x) essentially zero, confirming it is a root of this equation and not of any of the others.

Answer
x2+x−4=0x^2 + x - 4 = 0
Question 4
3 markshard
The equation x2−5=0x^2 - 5 = 0 can be rearranged to give the iterative formula xn+1=5xnx_{n+1} = \frac{5}{x_n}. Starting with x0=2x_0 = 2, describe what happens to the sequence of iterates.
Show worked solution

Worked solution

  1. Write down the formula and the starting value

    xn+1=5xn,x0=2x_{n+1} = \frac{5}{x_n}, \qquad x_0 = 2

    The rearrangement is valid algebra, but that alone does not make the iteration work.

  2. Work out x1x_{1}

    x1=2.5000x_{1} = 2.5000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  3. Work out x2x_{2}

    x2=2.0000x_{2} = 2.0000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  4. Work out x3x_{3}

    x3=2.5000x_{3} = 2.5000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  5. Work out x4x_{4}

    x4=2.0000x_{4} = 2.0000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  6. Work out x5x_{5}

    x5=2.5000x_{5} = 2.5000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  7. Work out x6x_{6}

    x6=2.0000x_{6} = 2.0000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  8. Look at what is happening

    x0=2.0000,  x1=2.5000,  x2=2.0000,  x3=2.5000x_0 = 2.0000,\; x_1 = 2.5000,\; x_2 = 2.0000,\; x_3 = 2.5000

    The terms simply repeat: x2=x0x_2 = x_0 and x3=x1x_3 = x_1.

  9. Explain why

    xn+2=5  5xn  =xnx_{n+2} = \frac{5}{\;\frac{5}{x_n}\;} = x_n

    Applying the formula twice returns you to where you started, so the sequence is locked into a two-value cycle and can never close in on the root 2.23612.2361.

  10. Conclude

    xn alternates: 2.0000,  2.5000,  2.0000,  2.5000,…x_n \text{ alternates: } 2.0000,\; 2.5000,\; 2.0000,\; 2.5000,\ldots

    The sequence never converges, so this iteration cannot be used to find the root.

Answer
The terms just flip back and forth between 2 and 2.5 for ever, so the sequence never converges and the iteration never gives the root.
Question 5
5 markschallenging
The equation x3−2x−5=0x^3 - 2x - 5 = 0 can be rearranged in more than one way. Iteration A: xn+1=xn3−52x_{n+1} = \frac{x_n^3 - 5}{2} with x0=2x_0 = 2. Iteration B: xn+1=2xn+53x_{n+1} = \sqrt[3]{2x_n + 5} with x0=2x_0 = 2. Which statement correctly describes what the two iterations do?
Show worked solution

Worked solution

  1. Both rearrangements are valid

    x3−2x−5=0  ⇒  x=x3−52andx=2x+53x^3 - 2x - 5 = 0 \;\Rightarrow\; x = \frac{x^3 - 5}{2} \quad\text{and}\quad x = \sqrt[3]{2x + 5}

    Each can be reversed to give the original equation, so both have the root as a fixed point. Validity is not the issue — behaviour is.

  2. Start iteration A

    xn+1=xn3−52,x0=2x_{n+1} = \frac{x_n^3 - 5}{2}, \qquad x_0 = 2

    Apply the formula and watch the terms.

  3. Iteration A: x1x_{1}

    x1=1.5000x_{1} = 1.5000

    Substitute the previous term into the right-hand side.

  4. Iteration A: x2x_{2}

    x2=−0.8125x_{2} = -0.8125

    Substitute the previous term into the right-hand side.

  5. Iteration A: x3x_{3}

    x3=−2.7682x_{3} = -2.7682

    Substitute the previous term into the right-hand side.

  6. Iteration A: x4x_{4}

    x4=−13.1061x_{4} = -13.1061

    Substitute the previous term into the right-hand side.

  7. Iteration A: x5x_{5}

    x5=−1128.1244x_{5} = -1128.1244

    Substitute the previous term into the right-hand side.

  8. Iteration A is diverging

    ∣x1∣=1.500→∣x5∣=1128.12|x_1| = 1.500 \to |x_5| = 1128.12

    The terms are flying away from the root, not towards it. Iteration A is useless for finding the root.

  9. Now start iteration B

    xn+1=2xn+53,x0=2x_{n+1} = \sqrt[3]{2x_n + 5}, \qquad x_0 = 2

    The same equation, a different rearrangement.

  10. Iteration B: x1x_{1}

    x1=2.080083…x_{1} = 2.080083\ldots

    Keep the full calculator display between steps.

  11. Iteration B: x2x_{2}

    x2=2.092350…x_{2} = 2.092350\ldots

    Keep the full calculator display between steps.

  12. Iteration B: x3x_{3}

    x3=2.094216…x_{3} = 2.094216\ldots

    Keep the full calculator display between steps.

  13. Iteration B: x4x_{4}

    x4=2.094500…x_{4} = 2.094500\ldots

    Keep the full calculator display between steps.

  14. Iteration B converges

    x4=2.09450  ⇒  x=2.095 (3 d.p.)x_4 = 2.09450 \;\Rightarrow\; x = 2.095 \text{ (3 d.p.)}

    The terms settle down on the root x=2.095x = 2.095 to 3 decimal places.

  15. Conclude

    A diverges,B→2.095\text{A diverges},\qquad \text{B} \to 2.095

    Two valid rearrangements of the same equation can behave completely differently; you must check that an iteration actually converges before trusting it.

Answer
A diverges; B converges to x=2.095 (3 d.p.)\text{A diverges; B converges to } x = 2.095 \text{ (3 d.p.)}

Unlock 65 more Iteration questions

Create a free account to work through every GCSE Iteration question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Iteration practice

Hard QuestionsWorked Solutions

Related Algebra topics

Algebraic fractionsEquation of a circleCoordinatesCubic and reciprocal graphsExponential and trig graphsExpanding binomials