He beat me to it

Earlier this week we worked on writing equations given the solutions to quadratics. I was surprised that a few were surprised that given solutions of 3\pm \sqrt{2} we can expand (x-(3+\sqrt{2}))(x-(3-\sqrt{2})) and get a “nice” equation. I reminded them that if we’re multiplying something with three terms by something with three terms we should get nine terms (before we simplify). After doing one more example but with complex roots, I sent them home with two new problems on which to practice. I was a bit worried as I had rushed this at the end of the period (my timing has been… off lately).

The next day, as they were comparing results and I was walking around observing, I noticed one student had this:

x=5 \pm \sqrt{3}

x -5=\pm \sqrt{3}

(x-5)^2=(\pm \sqrt{3})^2

x^2-10x+25=3

x^2-10x+22=0

So I asked him to put it up on the board and explain his steps. There was a collective “oooh” voiced as they watched him work.

I asked him how he thought of that. He said, “Isn’t that what you showed us yesterday?” After the class stopped laughing, he continued “Well, I didn’t actually write anything down that you did. So I just, uh, thought about it when I got home.”

However, the classes amazement quickly wore off, grumblings of “Why didn’t she show us this yesterday?” were heard. I replied, “Yesterday you wouldn’t have appreciated it. And I planned on showing you today, but he beat me to it.”

Have I mentioned that I really enjoy having an honors section?

Honestly, I just saw this method in my grad school class this semester. More than a few of the other math teachers in class had never seen it either, so I thought it was worth sharing.

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4 Responses to He beat me to it

  1. Kate Nowak says:

    I don’t know how to handle two posts in one day.

  2. Jackie says:

    Well, one was yesterday.

    Recently some asked if “Continuities” had been discontinued. That hurt more than I thought it would, so I figured it was time to start writing… something.

  3. Mr. H says:

    You can also turn this to a difference of squares.
    (x-(5+\sqrt{3}))(x-(5-\sqrt{3}))

    Distribute
    (x-5-\sqrt{3}))(x-5+\sqrt{3}))

    Regroup
    ((x-5)-\sqrt{3}))((x-5)+\sqrt{3}))

    If this were (a-b)(a+b), a=x-5 and b=\sqrt{3}, so we get a^2-b^2

    (x-5)^2-(\sqrt{3})^2

    [x^2-10x+25]-[3]

    x^2-10x+22

    This method also extends to complex solutions later on like (x-(2+3i))(x-(2-3i))

  4. cecil kirksey says:

    Hi:
    Your student is just working backwards from solving a quadratic equation by completing the square. I think.

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