Solving Equations with Roots

On Thursday we were finished with our self-guided projects on 5-9 and 5-10 and moved on the more math. Thursday also happened to be Pi day so we had some very good cookies and had some Pi related fun. We practiced reciting the digits from Pi, everyone did fairly well, and two students reached of 40 digits!

When we were done with our fun we moved on to learning about solving equations that had roots in them.

This little symbol is the square root sign. The solution then becomes whatever will square to become x. For example, the square root of 36 is 6, because 6^2 = 36.

Here is another image which displays the “index” part of roots. Whatever is in the index is the exponent that the solution would have to equal the radicand. In the example above, the solution is 3, because 3^3 = 27.

Here is an example of a problem where we simplified the square root of 32. 32 is not a perfect square, so we had to find a perfect square that was a factor of 32. 2 and 16 multiply to 32 so we can separate the two and then find the square root of 16, which is 4. So that means our simplified expression is 4*sqrt2.

In these four problems all you need to do is simplify within the root. In the third problem, we see -1^2 become 1, this is because when two negatives multiply, they become positive. When you have one term by itself that is squared, you can cancel it, but not if there are multiple terms. In problem one you can’t cancel because there is subtraction but in problem 4 you can cancel the square.

When you have fractions inside a root symbol, you can simplify the numerator and the denominator. We can see in problem 2 an example of solving with root 3. 3*3 equals 9, and 9*3 equals 27 so 27 becomes 3. 5*5 equals 25 and 25*5 equals 125 so 125 becomes 5. The simplified expression is 3/5.

Thanks for reading my blog! I hope you learned a lot!

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: