AQA · Mathematics · 8300
GCSE Maths: Iteration
Algebra · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Iteration (Corbettmaths Video 373), and every card links back to the exact page it came from. Practice tests are drawn from this deck’s own cards. Source material © Corbettmaths — used with attribution.
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How do you show that f(x) = 0 has a solution between x = a and x = b?
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Evaluate f(a) and f(b). If they have opposite signs (a change of sign), a root lies between a and b because a continuous function must cross zero.
Between which two consecutive integers does x³ − 8x − 10 = 0 have a solution?
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Between 3 and 4. f(3) = 27 − 24 − 10 = −7 and f(4) = 64 − 32 − 10 = 22, so there is a change of sign.
In an iterative formula xₙ₊₁ = g(xₙ), the starting value is written ?.
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x₀
What do the values x₁, x₂, x₃ produced by an iteration formula represent?
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Successive approximations to a root of the original equation, each usually closer to the true solution than the last.
What does it mean if x₁, x₂, x₃ from an iteration settle towards one fixed number?
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The iteration is converging: the values are approaching a root (solution) of the equation.
Rearrange x³ + 2x = 1 to make the x in the linear term the subject.
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x = (1 − x³) / 2, giving the iteration xₙ₊₁ = (1 − xₙ³) / 2.
Show that 3x − x³ = −11 has a solution between x = 2 and x = 3.
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Let f(x) = 3x − x³ + 11. f(2) = 9 and f(3) = −7; the change of sign means a root lies between 2 and 3.
Rearrange 3x − x³ = −11 into an iteration formula using a cube root.
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x³ = 3x + 11, so xₙ₊₁ = ∛(3xₙ + 11).
Why can one equation produce several different iteration formulae?
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It can be rearranged into x = g(x) in different ways; some rearrangements converge to a root while others diverge, so the choice of formula (and x₀) matters.
Show that x⁴ − 5x + 1 = 0 has a root between x = 1.5 and x = 2.
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f(1.5) = 5.0625 − 7.5 + 1 = −1.4375 and f(2) = 16 − 10 + 1 = 7; the change of sign means a root lies between 1.5 and 2.
To 2 decimal places, what is the root of x³ − 2x² + 19 = 0 in the interval (−3, −2)?
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x ≈ −2.14 (f(−2.14) ≈ 0.04 and f(−2.15) ≈ −0.18, so the root is −2.14 to 2 d.p.).
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