By Dawkins P.

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Extra resources for Algebra/Trig Review (2006)(en)(98s)

Example text

Solving Inequalities Solve each of the following inequalities. 1. x 2 − 10 > 3 x Solution To solve a polynomial inequality we get a zero on one side of the inequality, factor and then determine where the other side is zero. x 2 − 10 > 3 x ⇒ x 2 − 3 x − 10 > 0 ⇒ ( x − 5 )( x + 2 ) > 0 So, once we move everything over to the left side and factor we can see that the left side will be zero at x = 5 and x = −2 . These numbers are NOT solutions (since we only looking for values that will make the equation positive) but are useful to finding the actual solution.

I’ve got three versions of the inequality above. You can plug the test point into any of them, but it’s usually easiest to plug the test points into the factored form of the inequality. So, if you trust your factoring capabilities that’s the one to use. However, if you HAVE made a mistake in factoring, then you may end up with the incorrect solution if you use the factored form for testing. It’s a trade-off. The factored form is, in many cases, easier to work with, but if you’ve made a mistake in factoring you may get the incorrect solution.

In each region if the inequality is satisfied by one point from that region then it is satisfied for ALL points in that region. e it was positive at one point in region and negative at another) then it must also be zero somewhere in that region, but that can’t happen as we’ve already determined all the places where the polynomial can be zero! Likewise, if the inequality isn’t satisfied for some point in that region that it isn’t satisfied for ANY point in that region. e. small integers if possible) and plug it into the inequality.

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Algebra/Trig Review (2006)(en)(98s) by Dawkins P.

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