McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. Simplifying Radical Expressions
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Exercise 55 Page 249

Use the difference of squares when calculating the product.

See solution.

Practice makes perfect

We want to write two binomials of the given form and then find their product. asqrt(b)+ csqrt(f) and asqrt(b)- csqrt(f) Notice that these binomials contain square roots and, therefore, the radicands cannot be negative. This means that a and c can be any real numbers, but b and f need to be non-negative.

General Formula for the Product

We want to find the product of two binomials of the given form. Let's first find out how we can perform this multiplication in the general case, using the variables. To do so, we will use the difference of squares. ( asqrt(b)+ csqrt(f)) ( asqrt(b)- csqrt(f)) ⇓ ( asqrt(b))^2 - ( csqrt(f))^2 Next, we can simplify the obtained expression to get the general formula in the simplest form.

(asqrt(b))^2 - (csqrt(f))^2
a^2(sqrt(b))^2 - c^2(sqrt(f))^2
a^2b - c^2f

Example Binomials and Their Product

Now we can write two binomials of the given form. To do so, we will use a= 3, b= 5, c= 4, and f= 2. 3sqrt(5)+ 4sqrt(2) and 3sqrt(5)- 4sqrt(2) From here, we will find the product of these two binomials using the difference of squares.

(3sqrt(5)+4sqrt(2)) (3sqrt(5)-4sqrt(2))
(3sqrt(5))^2 - (4sqrt(2))^2
3^2(sqrt(5))^2 - 4^2(sqrt(2))^2
â–¼
Evaluate
9(sqrt(5))^2 - 16(sqrt(2))^2
9(5) - 16(2)
45-32
13

The product of our binomials is 13. Notice that when writing our binomials we could have substituted almost any real values we wanted. Therefore, our solution is only one of infinitely many possible solutions.

Using the General Formula

Finally, we can use our example binomials to check whether the general formula we have found is correct.

a^2b - c^2f
3^2( 5) - 4^2( 2)
â–¼
Evaluate
9(5)-16(2)
45-32
13

As we can see, our general formula gives the same result as the full multiplication.