Algebra II Essentials For Dummies (For Dummies (Math & by Mary Jane Sterling

By Mary Jane Sterling

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You now have to check to be sure that both your solutions work in the original equation. Remember: One or both may be extraneous solutions. Checking the original equation to see if the two solutions work, you first look at x = –1. Replace each x with –1: Nice! The first solution works. The next check is to see if is a solution. And right now I’m going to take “author’s privilege” and tell you that yes, the answer works. It takes more space than I have to show you all the steps, so I’m going to ask you to trust me and skip all the gory details.

In interval notation, you write the solution as (–6, 2]. Increasing the number of factors The method you use to solve a quadratic inequality (see the “Keeping it strictly quadratic” section, earlier in this chapter) works nicely with fractions and high-degree expressions. For example, you can solve (x + 2)(x – 4)(x + 7)(x – 5)2 ≥ 0 by creating a sign line and checking the products. The inequality is already factored, so you move to the step (Step 3) where you determine the zeros. The zeros are –2, 4, –7, and 5 (the 5 is a double root and the factor is always positive or 0).

Also, they factor in four different ways. When you have the proportion , the following are also true: ✓ ad and bc, the cross-products, are equal, giving you ad = bc. Chapter 4: Rolling Along with Rational and Radical Equations ✓ and , the reciprocals, are equal, giving 39 . ✓ You can divide out common factors both horizontally and vertically. Solve for x in the proportion: . First reduce across the numerators, and then reduce the left fraction: becomes becomes Now cross-multiply and solve the resulting quadratic equation: As usual, you need to check to be sure that you haven’t introduced any extraneous roots.

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