National 5MathsSQAFlashcardsAlgebraic SkillsCompleting the Square

Completing the Square (SQA National 5 Maths): Flashcards

Exam code: X847 75

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  • FrontCompleting the Square

    Complete the rule for rewriting x^{2} + p x with a squared bracket.

    x^{2} + p x = \left(x + \_\_\_\_\_\_\right)^{2} - \left(\_\_\_\_\_\_\right)^{2}

Cards in this collection (7)

  • Complete the rule for rewriting x^{2} + p x with a squared bracket.

    x^{2} + p x = \left(x + \_\_\_\_\_\_\right)^{2} - \left(\_\_\_\_\_\_\right)^{2}

    The completed rule is:

    x^{2} + p x = \left(x + \frac{p}{2}\right)^{2} - \left(\frac{p}{2}\right)^{2}

    Expanding the right-hand side confirms it, since \left(x + \frac{p}{2}\right)^{2} = x^{2} + p x + \left(\frac{p}{2}\right)^{2}.

  • Define the completed square form of a quadratic expression.

    The completed square form writes a quadratic as a squared bracket with a number added or subtracted, in the form \left(x + a\right)^{2} + b.

    For example x^{2} + 6 x - 11 has the completed square form \left(x + 3\right)^{2} - 20.

  • How is x^{2} + 10 x + 9 written in completed square form?

    Replace the first two terms using the rule, so x^{2} + 10 x becomes \left(x + 5\right)^{2} - 25.

    Bringing down the + 9 and simplifying the numbers gives \left(x + 5\right)^{2} - 16.

  • True or False?

    When x^{2} + p x is written with a squared bracket, a negative value of p changes the sign in front of the number at the end.

    False.

    The number taken away is \left(\frac{p}{2}\right)^{2}, which is a square and therefore never negative, so it is always subtracted.

    For example x^{2} - 20 x becomes \left(x - 10\right)^{2} - 100, with the 100 still taken away.

  • Fill in the two missing numbers to complete the square.

    x^{2} - 8 x + 5 = \left(x - \_\_\_\_\_\_\right)^{2} - \_\_\_\_\_\_

    The completed square form is:

    x^{2} - 8 x + 5 = \left(x - 4\right)^{2} - 11

    Squaring the bracket produces an extra 16 that has to be taken away again, and - 16 + 5 = - 11.

  • From y = \left(x + 3\right)^{2} - 12 how can you tell the smallest value y can take?

    A squared term can never be negative, so the smallest \left(x + 3\right)^{2} can be is 0.

    That makes the smallest possible value of y equal to - 12, so y \ge - 12.

  • How can you check a completed square without redoing the working?

    Expand the squared bracket and simplify, which should return the quadratic you started with.

    Expanding \left(x + 4\right)^{2} - 13 gives x^{2} + 8 x + 16 - 13, which is x^{2} + 8 x + 3.

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