Four ways of solving quadratic equations worked examples



When I teach solving quadratic equations I really like the pupils to have the opportunity to explore the pros and cons of each of the four solution methods. However, before they get there they need to be comfortable with all the methods… I like giving them a quadratic equation and then getting them to produce a poster showing all four ways of solving it. Giving them the following worked examples booklet helps many pupils in understanding how to calculate and present their solutions:


Download worked examples booklet

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3 Responses

  1. Thank you for the ideas. My comments:

    1. Solving graphically: Who on earth would go on to plot the graph when the table has already revealed the solutions?

    2. Solving by factorising: Is only working in nice cases, like the one given.

    3. The quadratic formula: OK, but where did the formula come from?

    4. Completing the square: OK, but quite a bit of work that may go wrong.

    Now the main question: are the students given time to find ways to solve quadratic equations (and problems that want them to solve quadratic equations!)?

    Here are some other ways:

    In other words, go to WolframAlpha and type in “solve y = x^2 + 2x – 8”.

    The line of symmetry is x = -b/(2a) = -1 so y(-1+d) = 0 for a d.
    Find d: (-1+d)^2 + 2(-1+d) – 8 = 0
    d^2-9=0 so d = 3 or d = -3, so the roots are 2 and -4.

    Trial and error.

    With trial and error find a and b where y(a) and y(b) have different signs and then chop the interval a-b in two till you have the root till a reasonable accuracy.

    How was it done historically without the use of the algebaric language? How did the Babylonians do it? How did Cardano do it?

    There must be thousands other ways. Why not give the students time to find some of them?

    • Here is another way that works for this simple example.
      Heuristic: Try to solve the problem with a method that works on related problems.

      x^2 + 2x – 8 = 0
      x^2 + 2x = 8
      x * (x + 2) = 8
      So we are looking for two numbers that are 2 apart and whose product is 8.
      2 * 4 = 8 so x = 2
      (-4) * (-2) = 8 so x = -4

      When you try this method on ax^2 + bx + c = 0 you get
      x * (ax + b) = -c which is not too nice.

      However, if you ‘solve for x’ you get
      x = (-c) / (ax + b)
      which may be written as a recursive function
      x(n+1) = (-c) / (ax(n) + b)
      and may reach the answer for some starting values x(0) which the students can experiment with on a spreadsheet or a simple program, e.g. in Python.

  2. Ian VanderSchee says:

    In answer to the question, “where did the quadratic formula come from,” if you solve “ax^2 + bx + c = 0” by completing the square, you get the Quadratic Formula.

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