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# Multiplying and Dividing With Square Roots

The following rules for multiplication and division with algebraic square roots are the same as those for numerical square roots:

• Property 1: • Property 2: where ‘a’ and ‘b’ stand for any valid mathematical expression.

Example 1:

Simplify solution:

We can simply go ahead and apply property (1), getting But now, we need to recall that simplification of square roots involves finding perfect square factors in the square root, so there is no point in expanding the product of the two binomials in this square root. Instead, we check for the possibility of further factoring

6x 2 + 24x = 6x(x + 4)

and

18x 4 + 72x 3 = 18x 3(x + 4)

Thus Example 2:

Expand and simplify solution:

At the start, this example involves the square of a binomial. You can use either of the two approaches presented earlier for multiplying one binomial by another binomial. The result here is as the final answer.

Two errors are commonly made with this type of problem:

(1) People sometimes just square each term of the binomial: But, you know that this is not the correct way to expand what amounts to the product of two binomials.

(2) People sometimes forget the obvious simplification that , leaving their answer as  Example 3:

Expand and simplify solution:

This example is very similar to the problem in Example 2 above – the product of two binomials, each of which contains a square root. We get  as the final result.

Example 4:

Expand and simplify solution: as the final result. The two middle terms in the second last line are identical but of opposite sign, and so cancel each other. This sort of pattern of binomial factors, where each binomial contains square roots but their product contains no square roots, is the basis for methods to rationalize denominators of fractions which have binomial denominators.

Example 5:

Simplify solution:

We can use either of two approaches here. We could start by using property (2) from the beginning of this document to write and then simplify the fraction in the square root to get as a final, simplified answer. This approach is quite short and efficient.

A second approach is to simplify the square roots first and then rationalize the denominator: But and so which is exactly the same final answer as we obtained with the first method.

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