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Welcome to GCSE Edexcel Maths revision.

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Topic M 20: Iteration and quadratic inequalities.

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This video covers Higher tier.

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Iteration repeatedly uses a formula to improve or generate values. x subscript open bracket n plus 1 close bracket is the next value calculated from the current value x subscript open bracket n close bracket ;

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x subscript open bracket 0 close bracket is the starting value.

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To solve x squared equals x plus 2,

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one possible iteration is x subscript open bracket n plus 1 close bracket equals square root of open bracket x subscript open bracket n close bracket plus 2 close bracket .

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A rearrangement is chosen, then the result is fed back as the next input.

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Worked example: With x subscript open bracket 0 close bracket equals 1,

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the successive values are x subscript open bracket 1 close bracket is approximately equal to 1.7321,

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x subscript open bracket 2 close bracket is approximately equal to 1.9319 and x subscript open bracket 3 close bracket is approximately equal to 1.9829;

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they approach the positive root 2.

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Keep full calculator precision between iterations.

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An iteration can converge, diverge or oscillate.

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A different rearrangement or starting value may behave differently, so do not assume repeated substitution always solves the equation.

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The square-root iteration above cannot find the negative root minus 1.

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Always consider the domain and whether other roots may exist.

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If asked for a value to a given number of decimal places,

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check that successive iterates stabilise at that precision;

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this is numerical evidence,

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not a guarantee for every iteration scheme.

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To locate a continuous function's root by a sign change, find values on opposite sides of zero.

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For f of open bracket x close bracket equals x cubed minus 2,

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f of open bracket 1 close bracket equals minus 1 and f of open bracket 2 close bracket equals 6,

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so at least one root lies between 1 and 2.

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Refine the bracket using trial values. f of open bracket 1.25 close bracket equals minus 0.046875 and f of open bracket 1.26 close bracket equals 0.000376,

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so the positive root lies between 1.25 and 1.26.

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State your interval and show the signs.

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Find the roots of the associated quadratic equation, then inspect where the graph is above or below the x-axis.

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The roots divide the number line into intervals.

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Negative part of x squared minus five x plus six

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Worked example: x squared minus 5 x plus 6 is less than 0 has roots 2 and 3.

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The upward-opening parabola is negative between the roots, so 2 is less than x is less than 3.

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For x squared minus 5 x plus 6 is greater than or equal to 0, the solution is x is less than or equal to 2 or x is greater than or equal to 3.

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There are two separate allowed intervals; joining them with and would be impossible.

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With is less than or equal to or is greater than or equal to include the roots; with is less than or is greater than exclude them.

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Use filled or open endpoints accordingly.

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If the leading coefficient is negative, the sign pattern reverses.

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For 6 minus x minus x squared is greater than 0, roots are minus 3 and 2, and the allowed interval is minus 3 is less than x is less than 2.

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Check one test value in each interval if unsure.

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Solving only the boundary equation does not answer an inequality question.

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That completes Iteration and quadratic inequalities.

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Revisit the notes and test yourself on the revision website.