quadratic formula
The Quadratic Formula
Part 2

Yesterday we began discussing the quadratic formula, and I asked you to solve a specific quadratic equation using the formula. You can see that post here: Solving Quadratic Equations with the Quadratic Formula

Today I will provide the solution to that problem. For your reference, let me give you the quadratic formula one more time:

The solutions to the quadratic equation ax2bx + c = 0 are

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Example: Solve the quadratic equation x2 – 2x – 150 by using the quadratic formula.


\frac{-b\pm \sqrt{b^2-4ac}}{2a}

=\frac{2\pm \sqrt{(-2)^2-4(1)(-15)}}{2\cdot 1}

=\frac{2\pm \sqrt{4+60}}{2}

=\frac{2\pm \sqrt{64}}{2}

=\frac{2\pm 8}{2}

=\frac{2}{2}\pm \frac{8}{2}

=1\pm 4

So, the two solutions are 1 + 4 = 5, and 1 – 4 = –3

Note: This particular problem could be solved more easily by factoring. 

Why does the quadratic formula look so messy? It’s not a very pleasant looking formula. It would have been nicer if we had a simpler formula for solving a quadratic equation. But unfortunately, this is simply what it turned out to be. As a challenging exercise, I would like you to see this for yourself firsthand. I will give a full solution tomorrow:

Challenge Problem: Solve the general quadratic equation ax2bx + c = 0 by completing the square, and note that this gives a derivation of the quadratic formula.

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