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The Quadratic Formula Explained (page 1 of 3) Often, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. So, while factoring may not always be successful, the Quadratic Formula can always find the solution.The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as:
Note that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root. And don't forget that it's a "2a" under there, not just a "2"! And make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. And remember that "b2" means "the square of ALL of b, including the sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive. In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this! Here are some examples of how the Quadratic Formula works:
Note first that this quadratic happens to factor: x2 + 3x – 4 = (x + 4)(x – 1) = 0 ...so x = –4 and x = 1. How would this look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, it looks like this:
Then, as expected, the solution is x = –4, x = 1. Recall that, when y = 0, you are finding the x-intercepts of the graph. So solving ax2 + bx + c = 0 for x means that, among other things, you are trying to find the x-intercepts. Since you came up with two solutions for this equation, there must be two x-intercepts on the graph. Graphing, you get the curve below:
As you can see, the x-intercepts match the solutions, falling at x = –4 and x = 1. This shows the connection between graphing and solving: When you are solving "(quadratic) = 0", you are finding the x-intercepts of the graph. This can be useful if you have a graphing calculator, because you can use the Quadratic Formula (when necessary) to solve a quadratic, and then use your graphing calculator to make sure that the displayed x-intercepts have the same decimal values as the solutions that the Quadratic Formula gives you. (Note that the calculator display on the graph will probably have some pixel-related round-off error, so you'd be checking to see that the values were close; don't expect them to be exact on the screen.) Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
Since there are no factors of (2)(–3) = –6 that add up to –4, then this quadratic cannot be factored. But I can always use the Quadratic Formula. In this case, a = 2, b = –4, and c = –3:
Then the answer is x = –0.58, x = 2.58, rounded to two decimal places. You would never get this solution by factoring! Note that the "solution" or "roots" or "zeroes" are usually required to be in the "exact" form of the answer (the one with the square roots of ten in it), but you'll need to get a calculator approximation in order to graph the x-intercepts or to simplify the final answer in a word problem. Unless you have a good reason to think that the answer is supposed to be a rounded answer, go with the exact form.
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Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
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![x = [ -b ± sqrt(b^2 - 4ac) ] / 2a](quads/qform01.gif){kind=small}
