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Solving Quadratic Equations: Examples (page 6 of 6)

Sections: Solving by: factoring, taking roots, completing the square, using the formula, graphing

The problem many students have is that they can solve quadratics when they're working on the homework, because the section that they're working on gives them hints. For instance, if you're working on the homework in the "Solving by Factoring" section, then you know that you're supposed to solve by factoring. But on the test, you don't know which section the quadratic came from. You could use the Quadratic Formula for everything, but the Formula can be "overkill". For example:

• Solve (x + 1)(x – 3) = 0.

This is a quadratic, and I'm supposed to solve it. I could multiply the left-hand side, find the coefficients, plug them into the Quadratic Formula, and chug away to the answer.

But why would I? I mean, for heaven's sake, this is factorable, and they've already set it equal to zero and factored it for me. So, while the Quadratic Formula would give me the correct answer, there is no reason to bother with it in this case. Instead, I'll just solve the factors:

(x + 1)(x – 3) = 0
x + 1 = 0  or  x – 3 = 0
x = –1  or  x = 3

The solution is  x = –1, 3.

• Solve x² + x – 4 = 0.

This one doesn't factor (since there are no factors of (1)(–4) = –4 that add to +1), and this isn't "squared-part equals a number", so I can't use square-rooting to solve. Then I can use completing the square (yuck!) or the Quadratic Formula. I'll use the Formula:

The solution is   .

• Solve x² – 3x – 4 = 0.

This one factors easily:

x² – 3x – 4 = 0
(x + 1)(x – 4) = 0
x + 1 = 0  or  x – 4 = 0
x = –1  or  x = 4

The solution is x = –1, 4.

• Solve x² – 4 = 0.

This is two terms, and nothing factors out, so either it's a difference of squares (so I can factor) or else it's "squared part equals a number" so I can square-root both sides. In this case, I can factor:

x² – 4 = 0
(x + 2)(x – 2) = 0
x + 2 = 0  or  x – 2 = 0
x = –2  or  x = 2

The solution is x = ± 2.

• Solve 6x² + 11x – 35 = 0.

This may factor, but it looks like it would be a fair amount of work, and I'm feeling mindless and lazy, so I'll use the Quadratic Formula instead:

The solutions are fractions, which means the quadratic could have been factored. But I've got my answer, so I don't care any more.

The solution is x = ^–7/₂, ⁵/₃.

• Solve x² – 48 = 0.

This is two terms, and nothing factors out, so either it's a difference of squares (so I can factor) or else it's "squared part equals a number" so I can square-root both sides. In this case, I have to square-root both sides:

The solution is  .

• Solve x² – 7x = 0.

This one factors easily: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

x² – 7x = 0
x(x – 7) = 0
x = 0  or  x – 7 = 0
x = 0  or  x = 7

The solution is x = 0, 7.

In general, you first check to see if there is an obvious factoring or if there is an obvious square-rooting that you can do. If not (that is, if there isn't a quick way of doing it), then you resort to the Quadratic Formula. The Quadratic Formula will always give you the answer, but in the simpler cases, use the other, faster methods.

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