The Quadratic Formula Explained

Transcription

1 The Quadratic Formula Explained Often, the simplest way to solve "ax 2 + 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. 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 "ax 2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients". The Formula is derived from the process of completing the square, and is formally stated as: For ax 2 + bx + c = 0, the value of x is given by: For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". 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. Remember that "b 2 " means "the square of ALL of b, including its sign", so don't leave b 2 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. Here are some examples of how the Quadratic Formula works: Solve x 2 + 3x 4 = 0 This quadratic happens to factor: x 2 + 3x 4 = (x + 4)(x 1) = 0...so I already know that the solutions are x = 4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = 4, my solution looks like this: 1

2 Then, as expected, the solution is x = 4, x = 1. Suppose you have ax 2 + bx + c = y, and you are told to plug zero in for y. The corresponding x- values are the x-intercepts of the graph. So solving ax 2 + bx + c = 0 for x means, among other things, that you are trying to find x-intercepts. Since there were two solutions for x 2 + 3x 4 = 0, there must then be two x-intercepts on the graph. Graphing, we get the curve below: As you can see, the x-intercepts match the solutions, crossing the x-axis 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 do the solutions provided by the Quadratic Formula. Note, however, that the calculator's display of the graph will probably have some pixel-related roundoff error, so you'd be checking to see that the computed and graphed values were reasonably close; don't expect an exact match. Copyright Elizabeth Stap All Rights Reserved Solve 2x 2 4x 3 = 0. Round your answer to two decimal places, if necessary. There are no factors of (2)( 3) = 6 that add up to 4, so I know that this quadratic cannot be factored. I will apply the Quadratic Formula. In this case, a = 2, b = 4, and c = 3: 2

3 Then the answer is x = 0.58, x = 2.58, rounded to two decimal places. Warning: The "solution" or "roots" or "zeroes" of a quadratic are usually required to be in the "exact" form of the answer. In the example above, the exact form is the one with the square roots of ten in it. 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. But unless you have a good reason to think that the answer is supposed to be a rounded answer, always go with the exact form. Compare the solutions of 2x 2 4x 3 = 0 with the x- intercepts of the graph: Remember: The "solutions" of an equation are also the x- intercepts of the corresponding graph. The Quadratic Formula: Solutions and the Discriminant Solve x(x 2) = 4. Round your answer to two decimal places. I not only cannot apply the Quadratic Formula at this point, I cannot factor either. I can not claim that "x = 4, x 2 = 4", because this is not how "solving by factoring" works. I must first rearrange the equation in the form "(quadratic) = 0", whether I'm factoring or using the Quadratic Formula. The first thing I have to do here is multiply through on the left-hand side, and then I'll move the 4 over: x(x 2) = 4 x 2 2x = 4 x 2 2x 4 = 0 Since there are no factors of (1)( 4) = 4 that add up to 2, then this quadratic does not factor. (In other words, there is no possible way that the faux-factoring solution of "x = 4, x 2 = 4" could be even slightly correct.) So factoring won't work, but I can use the Quadratic Formula; in this case, a = 1, b = 2, and c = 4: Copyright Elizabeth Stapel All Rights Reserved 3

4 Then the answer is: x = 1.24, x = 3.24, rounded to two places. For reference, here's what the graph looks like: There is a connection between the solutions from the Quadratic Formula and the graph of the parabola: you can tell how many x-intercepts you're going to have from the value inside the square root. The argument of the square root, the expression b 2 4ac, is called the "discriminant" because, by using its value, you can discriminate between (tell the differences between) the various solution types. Solve 9x x + 4 = 0. Using a = 9, b = 12, and c = 4, the Quadratic Formula gives: Then the answer is x = 2 / 3 4

5 In the previous examples, I had gotten two solutions because of the "plus-minus" part of the Formula. In this case, though, the square root reduced to zero, so the plus-minus didn't count for anything. This solution is called a "repeated" root, because x is equal to 2 / 3, but it's equal kind of twice: 2 / and 2 / 3 0. You can also see this repetition better if you factor: 9x x + 4 = (3x + 2)(3x + 2) = 0, so x = 2 / 3 and x = 2 / 3. Any time you get zero in the square root of the Quadratic Formula, you'll only get one solution. The square-root part of the Quadratic Formula is called "the discriminant", I suppose because you can use it to discriminate between whether the given quadratic has two solutions, one solution, or no solutions. This is what the graph looks like: The parabola only just touches the x-axis at x = 2 / 3 ; it doesn't actually cross. This is always true: if you have a root that appears exactly twice, then the graph will "kiss" the axis there, but not pass through. Solve 3x 2 + 4x + 2 = 0. Since there are no factors of (3)(2) = 6 that add up to 4, this quadratic does not factor. But the Quadratic Formula always works; in this case, a = 3, b = 4, and c = 2: At this point, I have a negative number inside the square root. If you haven't learned about complex numbers yet, then you would have to stop here, and the answer would be "no solution"; if you do know about complex numbers, then you can continue the calculations: If you do not know about complexes, then your answer would be "no solution". If you do know about complexes, then you would say there there is a "complex solution" and would give the answer (shown above) with the " i " in it. But whether or not you know about complexes, you know that you cannot 5

6 graph your answer, because you cannot graph the square root of a negative number. There are no such values on the x-axis. Since you can't find a graphable solution to the quadratic, then reasonably there should not be any x-intercept (because you can graph an x-intercept). Here's the graph: This relationship is always true: If you get a negative value inside the square root, then there will be no real number solution, and therefore no x-intercepts. (The relationship between the value inside the square root, the type of solutions, and the number of x-intercepts on the graph is summarized in a chart on the next page.) The Quadratic Formula: The Discriminant and Graphs Solve x 2 + 2x = 1. Round to two decimal places. I cannot apply the Quadratic Formula yet! The Formula only applies once I have "(quadratic) = 0", and I don't have that yet here. The first thing I have to do is move the 1 over, so I'll have "= 0" on the right-hand side: Copyright Elizabeth Stapel All Rights Reserved x 2 + 2x 1 = 0 Letting a = 1, b = 2, and c = 1, the Quadratic Formula gives me: Then the answer is x = 2.41, x = 0.41, rounded to two decimal places. 6

7 Here's the graph: The x-intercepts (that is, the solutions from above) are marked in red. This relationship between the value inside the square root (the discriminant), the type of solutions (two different solutions, one repeated solution, or no real solutions), and the number of x-intercepts (on the corresponding graph) of the quadratic is summarized in this table: x 2 2x 3 x 2 6x + 9 x 2 + 3x + 3 a positive number inside the square root zero inside the square root a negative number inside the square root two real solutions one (repeated) real solution two complex solutions 7

8 two distinct x-intercepts one (repeated) x- intercept no x-intercepts Probably the most important thing to remember when using the Quadratic Formula (other than the Formula itself, which you should memorize) is that you must do each step clearly and completely, so you don't lose your denominators or plus-minuses or square roots. Don't skip stuff, and you should do fine. Warning: If you get in the habit of "forgetting" the square root sign until the end when the back of the book "reminds" you that you "meant" to put it in, I'll bet good money that you'll mess up on your test. If you get in the habit of "forgetting" the plus/minus sign until the answer in the back "reminds" you that it belongs in there, then you will almost certainly miss every single problem where the answer doesn't have a square root symbol in it to "remind" you to put the plus/minus sign back in. That is, any time your answer is supposed to be something like "x = 5 ± 10", you will put down "x = = 15", and will have no idea how the book (or test) got the second answer of "x = 5". If you get sloppy with the denominator "2a", either by forgetting the "a" or by not dividing the entire numerator by this value, you will consistenly get the wrong answers. I've been grading homework and tests for too many years to be kidding about this. Really, truly; you want to do your work neatly and completely every single time! 8