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Beginning and Intermediate Algebra - Wallace Math Courses ...

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Using i we can simplify radicals with negatives under the root. We will use the<br />

product rule <strong>and</strong> simplify the negative as a factor of negative one. This is shown<br />

in the following examples.<br />

Example 438.<br />

√<br />

− 16<br />

√<br />

− 1 · 16<br />

Example 439.<br />

√<br />

− 24<br />

√<br />

− 1 · 4 · 6<br />

2i 6<br />

√<br />

Consider the negative asafactor of − 1<br />

Take each root, square root of − 1 isi<br />

4i Our Solution<br />

Find perfect square factors, including − 1<br />

Square root of − 1 isi, square root of 4 is 2<br />

Our Solution<br />

When simplifying complex radicals it is important that we take the − 1 out of the<br />

radical (as an i) before we combine radicals.<br />

Example 440.<br />

√ √<br />

− 6 − 3 Simplify the negatives, bringing i out of radicals<br />

(i 6<br />

√ )(i 3<br />

√ ) Multiply i by i isi2 = − 1, also multiply radicals<br />

√<br />

− 18 Simplify the radical<br />

√<br />

− 9 · 2 Take square root of9<br />

− 3 2<br />

√<br />

Our Solution<br />

If there are fractions, we need to make sure to reduce each term by the same<br />

number. This is shown in the following example.<br />

Example 441.<br />

√<br />

− 15 − − 200<br />

Simplify the radical first<br />

20 √<br />

− 200 Find perfect square factors, including − 1<br />

√<br />

− 1 · 100 · 2 Take square root of − 1 <strong>and</strong> 100<br />

10i 2<br />

√<br />

Put this back into the expression<br />

− 15 − 10i 2<br />

√<br />

All the factors are divisible by 5<br />

20<br />

− 3 − 2i 2<br />

√<br />

Our Solution<br />

4<br />

√<br />

By using i = − 1 we will be able to simplify <strong>and</strong> solve problems that we could<br />

not simplify <strong>and</strong> solve before. This will be explored in more detail in a later section.<br />

322

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