Adding and Subtracting Radicals

DAY 1   

On Wednesday, March 20, we worked with square roots again. However, instead of multiplying them we added the roots. At the beginning of class we were asked to solve four problems on the board. The first was

We solved the first problem by factoring and then combining common terms. 

The next problem was a bit more difficult because it contained division. Also there were two different methods of solving it.   Here are the two methods:

                                     

In the third problem we had to use the foil method and learned that you can only multiply out numbers with the same index. 

Example of multiplying roots with (1) the same index and (2) different indices.

Example of solving a problem using the foil method when the numbers have different index numbers.

The final problem introduced variables and fractions into the mix. Treat the variables like any other number. To deal with the fractions, you must make all the denominators the same. The easiest way to do this is to find the least common denominator, or the smallest whole number that is divisible by all the denominators in the problem. This is the same idea behind finding the least common multiple for whole numbers. Once you find the least common denominator, multiply the original denominator by the least common denominator divided by that original denominator (New Denominator = LCD). Then multiply the numerator of the preceding fraction by the LCD divided by the denominator (New Numerator = (LCD ÷ Denominator) x Numerator.

 For example

However,  if this question were in a textbook, the preceding answer would not be acceptable because it is not rationalized. Rationalizing is making sure that there is no root in the denominator. To rationalize our answer,  you must multiply the numerator and the denominator by the denominator. For example

After learning all the above, the class put our knowledge to the test by solving the following more challenging problem:

If there is a regular octagon inscribed in a square that has lengths of 4, find the length of one of the octagon's sides.

   

To solve the problem take the following steps:

1. Label as x the legs of the isosceles triangles found in each corner of the square.                                    

2. Since the isosceles triangle are right triangles, label the hypotenuses √2x^2.

3. Given that each side of a square is 4 and the sum of the lengths of the legs of the two isosceles triangles that make up part of that side of the square are equal to 2x, 4-2x would equal a side of the octagon.

4. If the octagon is regular then √2x^2 = 4 - 2x.

5. √2x^2 = x√2

6. x√2 + 2x = 4

7. x(√2 + 2) = 4

8. x = 4/(√2 + 2) = 1.17

9. The length of each side of the octagon is 2.7378.

Although this answer is correct, there is another possible answer. To find this we must use quadratics. The quadratic formula is a way to factor quadratics.   

Quadratic Formula

In the quadratic formula, a is the coefficient of x^2, b is coefficient of x, and c is the number that is not multiplied by a variable. For example

Now that we have reviewed quadratics back to our problem of determining the sides of the octagon.

1. If the octagon is regular then √2x^2 = 4 - 2x.

2. 2x^2 = (4 - 2x)(4 - 2x)

3. 2x^2 = 4x^2 - 16x + 16 

4. 0 = 2x^2 - 16x + 16

5. 0 = x^2 - 8x +8

6. x = (8 - √64 - 4 (1)(8))/2 or  x = (8 = √64 - 4 (1)(8))/2  (using quadratic formula)

7. x = (8 - √32)/2  or x = (8 + √32)/2

8. x = (8 - 4√2)/2 or x = (8 + 4√2)/2  

9. x = 4 - 2√2 or x = 4 + 2√2

10. x = 6.82 or x = 1.17

11. If you substitute 6.82 in the equation 4-2x, the side length is negative, which is impossible. If you substitute 6.82 in the equation √2x^2, the side length is 9.64, which makes the octagon larger than the square it is inscribed in. Therefore, x needs to equal 1.17.

11. The length of each side equals 2.7378.

Here is a video that reviews the basics of adding and subtracting radicals: