Hard GCSE Iteration Questions

Challenging, exam-style GCSE Iteration questions with worked solutions. Stretch yourself on the hardest sign change, iterative formula, repeated substitution, rounding problems.

sign changeiterative formularepeated substitutionroundingconvergenceroot to a given accuracy
GCSE Higher34 questionsStep-by-step solutions
Question 1
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.)}
Question 2
5 markschallenging
The equation x3+x−5=0x^3 + x - 5 = 0 can be rearranged in more than one way. Iteration A: xn+1=5−xn3x_{n+1} = 5 - x_n^3 with x0=1x_0 = 1. Iteration B: xn+1=5−xn3x_{n+1} = \sqrt[3]{5 - x_n} with x0=1x_0 = 1. Which statement correctly describes what the two iterations do?
Show worked solution

Worked solution

  1. Both rearrangements are valid

    x3+x−5=0  ⇒  x=5−x3andx=5−x3x^3 + x - 5 = 0 \;\Rightarrow\; x = 5 - x^3 \quad\text{and}\quad x = \sqrt[3]{5 - x}

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

    Apply the formula and watch the terms.

  3. Iteration A: x1x_{1}

    x1=4.0000x_{1} = 4.0000

    Substitute the previous term into the right-hand side.

  4. Iteration A: x2x_{2}

    x2=−59.0000x_{2} = -59.0000

    Substitute the previous term into the right-hand side.

  5. Iteration A: x3x_{3}

    x3=205384.0000x_{3} = 205384.0000

    Substitute the previous term into the right-hand side.

  6. Iteration A: x4x_{4}

    x4=−8.66363×1015x_{4} = -8.66363 \times 10^{15}

    Substitute the previous term into the right-hand side.

  7. Iteration A: x5x_{5}

    x5=6.50279×1047x_{5} = 6.50279 \times 10^{47}

    Substitute the previous term into the right-hand side.

  8. Iteration A is diverging

    ∣x1∣=4.000→∣x5∣=6.50279×1047|x_1| = 4.000 \to |x_5| = 6.50279 \times 10^{47}

    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=5−xn3,x0=1x_{n+1} = \sqrt[3]{5 - x_n}, \qquad x_0 = 1

    The same equation, a different rearrangement.

  10. Iteration B: x1x_{1}

    x1=1.587401…x_{1} = 1.587401\ldots

    Keep the full calculator display between steps.

  11. Iteration B: x2x_{2}

    x2=1.505549…x_{2} = 1.505549\ldots

    Keep the full calculator display between steps.

  12. Iteration B: x3x_{3}

    x3=1.517491…x_{3} = 1.517491\ldots

    Keep the full calculator display between steps.

  13. Iteration B: x4x_{4}

    x4=1.515760…x_{4} = 1.515760\ldots

    Keep the full calculator display between steps.

  14. Iteration B converges

    x4=1.51576  ⇒  x=1.516 (3 d.p.)x_4 = 1.51576 \;\Rightarrow\; x = 1.516 \text{ (3 d.p.)}

    The terms settle down on the root x=1.516x = 1.516 to 3 decimal places.

  15. Conclude

    A diverges,B→1.516\text{A diverges},\qquad \text{B} \to 1.516

    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=1.516 (3 d.p.)\text{A diverges; B converges to } x = 1.516 \text{ (3 d.p.)}
Question 3
5 markschallenging
The iterative formula xn+1=70xn+2x_{n+1} = \sqrt{\frac{70}{x_n + 2}}, started at x0=3x_0 = 3, converges to a number xx. This number is a root of which of these equations?
Show worked solution

Worked solution

  1. Write down the equation satisfied by the limit

    x=70x+2x = \sqrt{\frac{70}{x + 2}}

    At the limit, putting xx in gives xx back out.

  2. Clear the root or fraction

    x2=70x+2x^2 = \frac{70}{x + 2}

    Square both sides.

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

    x2(x+2)=70  ⇒  x3+2x2−70=0x^2(x + 2) = 70 \;\Rightarrow\; x^3 + 2x^2 - 70 = 0

    Multiply both sides by (x+2)(x + 2), expand, and take the 70 across.

  4. State the equation

    x3+2x2−70=0x^3 + 2x^2 - 70 = 0

    This is the equation whose root the iteration finds.

  5. Run the iteration to check

    x1=703+2=3.741657…x_1 = \sqrt{\frac{70}{3 + 2}} = 3.741657\ldots

    Start from x0=3x_0 = 3.

  6. Continue the iteration

    x2=70x1+2=3.491647…x_2 = \sqrt{\frac{70}{x_{1} + 2}} = 3.491647\ldots

    Keep the full calculator value of x1x_1.

  7. Continue the iteration

    x3=70x2+2=3.570242…x_3 = \sqrt{\frac{70}{x_{2} + 2}} = 3.570242\ldots

    Keep the full calculator value of x2x_2.

  8. Continue the iteration

    x4=70x3+2=3.544965…x_4 = \sqrt{\frac{70}{x_{3} + 2}} = 3.544965\ldots

    The values are settling down.

  9. The limit

    x5=3.553036…x_5 = 3.553036\ldots

    The iteration has converged.

  10. Substitute the limit into the equation

    f(3.553036)≈0.101886f(3.553036) \approx 0.101886

    The limit makes f(x)f(x) essentially zero, so it really is a root of this equation.

  11. Check the sign change

    f(3)=−25,f(4)=26f(3) = -25,\quad f(4) = 26

    The root lies between 33 and 44, and the iterates do too.

  12. Test one of the other options

    x3+2x2−7=0  ⇒  x=7x+2x^3 + 2x^2 - 7 = 0 \;\Rightarrow\; x = \sqrt{\frac{7}{x + 2}}

    This equation gives 7/(x+2)\sqrt{7/(x+2)}, not 70/(x+2)\sqrt{70/(x+2)} — the constant is wrong.

  13. Why the wrong options fail

    x3−2x2−70=0  ⇒  x=70x−2x^3 - 2x^2 - 70 = 0 \;\Rightarrow\; x = \sqrt{\frac{70}{x - 2}}

    A sign slip inside the bracket changes the formula, so this equation is not the one being solved either. Only x3+2x2−70=0x^3 + 2x^2 - 70 = 0 reverses back to the given iteration.

  14. The rearrangement must be reversible

    x=70x+2  ⟺  x3+2x2−70=0x = \sqrt{\frac{70}{x + 2}} \;\Longleftrightarrow\; x^3 + 2x^2 - 70 = 0

    Every step used to get from the equation to the formula can be undone, so the fixed point of the iteration is exactly a root of the equation.

  15. State the answer

    x3+2x2−70=0x^3 + 2x^2 - 70 = 0

    The iteration converges to a root of this equation and of no other option.

Answer
x3+2x2−70=0x^3 + 2x^2 - 70 = 0
Question 4
5 markschallenging
f(x)=x3+x−5f(x) = x^3 + x - 5. The iteration xn+1=5−xn3x_{n+1} = \sqrt[3]{5 - x_n} with x0=1x_0 = 1 gives x5=1.51601x_5 = 1.51601 and x6=1.51598x_6 = 1.51598. Which argument correctly shows that the root of f(x)=0f(x) = 0 is 1.5161.516 correct to 3 decimal places?
Show worked solution

Worked solution

  1. Note the candidate root

    x5=1.51601,x6=1.51598x_5 = 1.51601,\quad x_6 = 1.51598

    Both iterates round to 1.5161.516, so 1.5161.516 is the candidate — but agreement between iterates is not a proof of accuracy.

  2. Say what "1.516 to 3 d.p." means

    1.5155≤x<1.51651.5155 \le x < 1.5165

    These are exactly the numbers that round to 1.5161.516 at 3 decimal places.

  3. State the lower bound of the rounding interval

    xlow=1.5155x_{\text{low}} = 1.5155

    Half a unit of the last decimal place below the candidate.

  4. Substitute x=1.5155x = 1.5155

    f(1.5155)=1.51553+1.5155−5f(1.5155) = 1.5155^{3} + 1.5155 - 5

    Substitute the lower bound into f(x)f(x).

  5. Evaluate f(1.5155)f(1.5155)

    f(1.5155)=−0.003790151125f(1.5155) = -0.003790151125

    This is negative.

  6. Substitute x=1.5165x = 1.5165

    f(1.5165)=1.51653+1.5165−5f(1.5165) = 1.5165^{3} + 1.5165 - 5

    Now the upper bound.

  7. Evaluate f(1.5165)f(1.5165)

    f(1.5165)=0.004104617125f(1.5165) = 0.004104617125

    This is positive.

  8. Compare the signs

    f(1.5155)=−0.003790151125,f(1.5165)=0.004104617125f(1.5155) = -0.003790151125,\qquad f(1.5165) = 0.004104617125

    The signs are opposite: there is a change of sign across the interval.

  9. Conclude where the root is

    1.5155<x<1.51651.5155 < x < 1.5165

    ff is a polynomial and therefore continuous, so it crosses zero inside the interval.

  10. Every number in the interval rounds the same way

    1.5155≤x<1.5165  ⇒  x=1.516 (3 d.p.)1.5155 \le x < 1.5165 \;\Rightarrow\; x = 1.516 \text{ (3 d.p.)}

    So the root is 1.5161.516 to 3 decimal places.

  11. Why f(1.516)≈0f(1.516) \approx 0 is not enough

    f(1.516)=0.000156096f(1.516) = 0.000156096

    f(1.516)f(1.516) being small tells you 1.5161.516 is near a root, but not that the root rounds to 1.5161.516. Only the sign change over the rounding interval proves that.

  12. Why a wider interval is not enough

    1.515<x<1.5171.515 < x < 1.517

    A sign change over this wider interval would allow roots that round to 1.515 or 1.517 as well — it proves nothing about the 3th decimal place.

  13. Check against the iteration

    x6=1.515975615…x_6 = 1.515975615\ldots

    The unrounded iterate sits comfortably inside the rounding interval, as it must.

  14. Confirm the iterate lies in the interval

    1.5155<1.515976<1.51651.5155 < 1.515976 < 1.5165

    Everything is consistent.

  15. State the justified root

    f(1.5155)=−0.003790151125,  f(1.5165)=0.004104617125⇒x=1.516 (3 d.p.)f(1.5155) = -0.003790151125,\; f(1.5165) = 0.004104617125 \Rightarrow x = 1.516 \text{ (3 d.p.)}

    Quote both values, point out the change of sign, and state the conclusion — that is the full mark scheme.

Answer
f(1.5155)=−0.003790151125 and f(1.5165)=0.004104617125:sign change⇒x=1.516 (3 d.p.)f(1.5155) = -0.003790151125 \text{ and } f(1.5165) = 0.004104617125: \text{sign change} \Rightarrow x = 1.516 \text{ (3 d.p.)}
Question 5
5 markschallenging
f(x)=x3−2x−5f(x) = x^3 - 2x - 5. The iteration xn+1=2xn+53x_{n+1} = \sqrt[3]{2x_n + 5} with x0=2x_0 = 2 gives x5=2.09454x_5 = 2.09454 and x6=2.09455x_6 = 2.09455. Which argument correctly shows that the root of f(x)=0f(x) = 0 is 2.0952.095 correct to 3 decimal places?
Show worked solution

Worked solution

  1. Note the candidate root

    x5=2.09454,x6=2.09455x_5 = 2.09454,\quad x_6 = 2.09455

    Both iterates round to 2.0952.095, so 2.0952.095 is the candidate — but agreement between iterates is not a proof of accuracy.

  2. Say what "2.095 to 3 d.p." means

    2.0945≤x<2.09552.0945 \le x < 2.0955

    These are exactly the numbers that round to 2.0952.095 at 3 decimal places.

  3. State the lower bound of the rounding interval

    xlow=2.0945x_{\text{low}} = 2.0945

    Half a unit of the last decimal place below the candidate.

  4. Substitute x=2.0945x = 2.0945

    f(2.0945)=2.09453−2×2.0945−5f(2.0945) = 2.0945^{3} - 2 \times 2.0945 - 5

    Substitute the lower bound into f(x)f(x).

  5. Evaluate f(2.0945)f(2.0945)

    f(2.0945)=−0.000574591375f(2.0945) = -0.000574591375

    This is negative.

  6. Substitute x=2.0955x = 2.0955

    f(2.0955)=2.09553−2×2.0955−5f(2.0955) = 2.0955^{3} - 2 \times 2.0955 - 5

    Now the upper bound.

  7. Evaluate f(2.0955)f(2.0955)

    f(2.0955)=0.010592483875f(2.0955) = 0.010592483875

    This is positive.

  8. Compare the signs

    f(2.0945)=−0.000574591375,f(2.0955)=0.010592483875f(2.0945) = -0.000574591375,\qquad f(2.0955) = 0.010592483875

    The signs are opposite: there is a change of sign across the interval.

  9. Conclude where the root is

    2.0945<x<2.09552.0945 < x < 2.0955

    ff is a polynomial and therefore continuous, so it crosses zero inside the interval.

  10. Every number in the interval rounds the same way

    2.0945≤x<2.0955  ⇒  x=2.095 (3 d.p.)2.0945 \le x < 2.0955 \;\Rightarrow\; x = 2.095 \text{ (3 d.p.)}

    So the root is 2.0952.095 to 3 decimal places.

  11. Why f(2.095)≈0f(2.095) \approx 0 is not enough

    f(2.095)=0.005007375f(2.095) = 0.005007375

    f(2.095)f(2.095) being small tells you 2.0952.095 is near a root, but not that the root rounds to 2.0952.095. Only the sign change over the rounding interval proves that.

  12. Why a wider interval is not enough

    2.094<x<2.0962.094 < x < 2.096

    A sign change over this wider interval would allow roots that round to 2.094 or 2.096 as well — it proves nothing about the 3th decimal place.

  13. Check against the iteration

    x6=2.094550307…x_6 = 2.094550307\ldots

    The unrounded iterate sits comfortably inside the rounding interval, as it must.

  14. Confirm the iterate lies in the interval

    2.0945<2.094550<2.09552.0945 < 2.094550 < 2.0955

    Everything is consistent.

  15. State the justified root

    f(2.0945)=−0.000574591375,  f(2.0955)=0.010592483875⇒x=2.095 (3 d.p.)f(2.0945) = -0.000574591375,\; f(2.0955) = 0.010592483875 \Rightarrow x = 2.095 \text{ (3 d.p.)}

    Quote both values, point out the change of sign, and state the conclusion — that is the full mark scheme.

Answer
f(2.0945)=−0.000574591375 and f(2.0955)=0.010592483875:sign change⇒x=2.095 (3 d.p.)f(2.0945) = -0.000574591375 \text{ and } f(2.0955) = 0.010592483875: \text{sign change} \Rightarrow x = 2.095 \text{ (3 d.p.)}

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