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Proving the Quadratic Formula

Lesson Plans > Mathematics > Algebra > Factoring > Quadratic Formula
 

Proving the Quadratic Formula

One of the things I enjoy doing with my Algebra Two classes is "sneaking up" on a proof of the quadratic formula, without warning them that that's my objective. It's fun to watch the expressions of surprise on their faces, or the quiet gasps or chuckles, when they realize where the lesson is headed. Here's how the lesson goes. Keep in mind that in a previous lesson I've already introduced the quadratic formula, and shown how it works.

  1. Explain what completing the square is. Demonstrate this process by showing that if you have an expression like x2 + 8x + ___, and you want to fill in the blank to make a perfect square trinomial, you divide the coefficient of x by 2, and then square it. Thus, x2 + 8x + 16 is a perfect square trinomial. I give several examples. Some we do together on the board, and some they work on their own. Here are some examples:

    x2 - 6x + ____ (add 9)
    x2 - 12x + ____ (add 36)
    x2 + x + ____ (add
    1
    4
    )
    x2 + kx + ____ (add
    k2
    4
    )
     
  2. Remind them of one of the fundamental rules of manipulating equations - that if you do something to one side, you have to do it to the other side as well. In other words, if we have 3x - 5 = 0, can we add 5 to one side without adding it to the other? Students are quick to agree that you can't add something to one side unless you add it to the other as well. I then tell them they're wrong, and show them that I can add 5 to one side, without adding it to the other, as long as I subtract it right back out again:

    3x - 5 + 5 - 5 = 0

    Typically students think this is pointless and silly, and I tell them that normally it would be, but sometimes it works like a magic trick, and helps us to do things we normally couldn't do.
     
  3. Now I show them how completing the square and the add-subtract trick can be used to solve a quadratic equation:

    x2 + 14x + 48 = 0
    x2 + 14x + ____ - ______ + 48 = 0
    x2 + 14x + 49 - 49 + 48 = 0
    (x + 7)2 - (49 - 48) = 0
    (x + 7)2 - 1 = 0
    (x + 7)2 - 12 = 0
    (x + 7 + 1)(x + 7 - 1) = 0
    (x + 8)(x + 6) = 0
    x = -8 or  x = -6 

    I give them several problems to solve like this and then point out to them that we haven't done anything ground-breaking here; every problem we've done could also have been done by one of our other factoring tactics. 
     
  4. Now we break new ground by factoring something that doesn't come out as nice: x2 + 4x + 1 = 0

    x2 + 4x + ___ - ___ + 1 = 0
    x2 + 4 x+ 4 - 4 + 1 =0
    (x + 2)2 - 3 = 0
    (x + 2)2 -
    3
    2 = 0
    (x + 2 +
    3
    )(x + 2 -
    3
    ) = 0
    x = -2 -
    3
    or x = - 2 +
    3


    We go through several examples like this.
     
  5. Now I suggest that we raise the stakes a bit by sticking an extra variable in there, like this: x2 + 6x + k = 0

    x2 + 6x + ____ - ____ + k = 0
    x2 + 6x + 9 - 9 + k = 0
    (x + 3)2 - (9 - k) = 0
    (x + 3)2 -
    9 - k
    2 = 0
    (x + 3 -
    9 - k
    )(x + 3 +
    9 - k
    ) = 0
    x = -3 +
    9 - k
    or x = -3 -
    9 - k

     
  6. I then do another example in which we replace the coefficient of x with a variable: x2 + mx + 2 = 0

    x2 + mx + ____ - _____ + 2 = 0
    x2 + mx +
    m2
    4
    -
    m2
    4
    +  2 = 0
    (x +
    m
    2
    )2 - (
    m2
    4
    - 2) = 0
    (x +
    m
    2
    -
    m2
    4
    + 2
    )(x +
    m
    2
    +
    m2
    4
    + 2
    ) = 0
     
  7. After doing all of this, I put this up on the board: ax2 + bx + c = 0. Then I say, "Wait a minute, there's an a out in front! We've never done one like this before. Can we get rid of that a somehow?" Someone will usually suggest dividing the whole equation by a:

    x2 +
    b
    a
    x +
    c
    a
    = 0
    x2 +
    b
    a
    x + ____ - ____ +
    c
    a
    = 0
    x2 +
    b
    a
    x +
    b2
    4a2
    -
    b2
    4a2
    +
    c
    a
    = 0 
    (x +
    b
    2a
    )2 - (
    b2
    4a2
    -
    c
    a
    ) = 0
    (x +
    b
    2a
    )2 - (
    b2
    4a2
    -
    c
    a
    )2 = 0
    (x +
    b
    2a
    -
    b2
    4a2
    -
    c
    a
    )(x +
    b
    2a
    +
    b2
    4a2
    -
    c
    a
    ) = 0

    The next step is where some people start to recognize where we're headed, because I ask, "Can we simplify that ugly thing under the radical?" The answer is, YES! By finding a common denominator for the two fractions inside the radical:

    b2
    4a2
    -
    4ac
    4a2
    =
    b2 - 4ac
    4a2
    =
    b2 - 4ac
    2a


    That simplification process lets them see that infamous expression
    b2 - 4ac
    , which the're already familiar with, and  many will suddenly realize where we're headed. From here it's a mere matter of a little bit of simplification to obtain:

    x =
    -b ±
    b2 - 4ac
    2a
Lesson by Mr. Twitchell

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