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The Quadratic Formula:
     Solutions and the Discriminant (page 2 of 3)

• 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.

Do not try to say "x = 4, x – 2 = 4". This is not how it works! You must have "(quadratic) = 0" first, whether you're 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
x² – 2x = 4
x² – 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 ever be even slightly correct.) I can use the Quadratic Formula; in this case, a = 1,
b = –2, and c = –4:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

Then the answer is:  x = –1.24, x = 3.24, rounded to two places.

There is a connection between the solutions from the Quadratic Formula and the graph of the parabola. In particular, you can tell how many x-intercepts you're going to have from the value inside the square root.

• Solve 9x² + 12x + 4 = 0.

Using a = 9, b = 12, and c = 4, the Quadratic Formula gives:

Then the answer is x = ^–2/₃.

In the previous cases, I got two solutions because of the "plus-minus" part. 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: ^–2/₃ + 0 and ^–2/₃ – 0. You can also see this if you factor:  9x² + 12x + 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.

Note that the parabola 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 3x² + 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". But whether or not you know about complexes, you know that you cannot graph your answer, because you cannot graph the square root of a negative number. Since you can't find a graphable solution, then reasonably there should not be any x-intercept (because you can graph an x-intercept). 

This relationship is always true: If you get a negative inside the square root, then there will be no real number solution, and therefore no x-intercepts.

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