Solving Radical Equations - Extraneous Roots

Category: Algebra

"Published in Newark, California, USA

Find the roots of the equation
 

Solution:

Consider the given equation
 

Square both sides of the equation
 

 

 

 

Square both sides of the equation again
 

 

 

 

Equate each factor to zero to solve for the value of x

For: x - 9 = 0                                                      For: x - 2 = 0
       x = 9                                                                  x = 2

Now, check the values of x to the given equation

If x = 9
 

 

 

 

                                         x = 9 is an extraneous root.

If x = 2
 

 

 

 

 

                                          x = 2 is a root.

Therefore, 2 is the root.

Venn Diagram Problem

Category: Algebra

"Published in Newark, California, USA"

A school has 63 students who will take Physics, Chemistry and Biology. 33 students are taking Physics, 25 students are taking Chemistry, and 26 students are taking Biology. 10 students are taking Physics and Chemistry, 9 students are taking Biology and Chemistry, and 8 students are taking Physics and Biology. If a number of students are taking all three subjects is the same as a number of students are not taking either of the subjects, how many students are taking all the three subjects? How many students are taking only one of the three subjects?

Solution:

From the given problem above, it is a Venn Diagram Problem because it involves the intersection or mutual items of the sets. Consider the figure below:

Photo by Math Principles in Everyday Life

Let x be the number of students are taking Physics, Chemistry and Biology

By further analysis,

Number of students are taking Physics and Chemistry alone = 10 - x

Number of students are taking Physics and Biology alone = 8 - x

Number of students are taking Chemistry and Biology alone = 9 - x

Number of students are taking Physics alone = 33 - (10 - x) - x - (8 - x) = 33 - 10 + x - x - 8 + x = 15 + x

Number of students are taking Chemistry alone = 25 - (10 - x) - x - (9 - x) = 25 - 10 + x - x - 9 + x = 6 + x

Number of students are taking Biology alone = 26 - (8 - x) - x - (9 - x) = 26 - 8 + x - x - 9 + x = 9 + x

Next, we put all the items and parameters in the figure, we have

Photo by Math Principles in Everyday Life

Referring to the problem statement, "if a number of students are taking all three subjects is the same as a number of students are not taking either of the subjects


(15 + x) + (10 - x) + x + (8 - x) + (6 + x) + (9 - x) + (9 + x ) = 63 - x

(Let 63 - x be the number of students are taking Physics, Chemistry, and Biology. 63 is the total number of students in a school. x is the number of students are not taking either of the subjects.)

                             57 + x = 63 - x

                               x + x = 63 - 57

                                   2x = 6

                                     x = 3

Therefore, there are 3 students taking Physics, Chemistry, and Biology. 

Number of students are taking Physics alone = 15 + x = 15 + 3 = 18

Number of students are taking Chemistry alone = 6 + x = 6 + 3 = 9

Number of students are taking Biology alone = 9 + x = 9 + 3 = 12

Therefore, there are 18 + 9 + 12 = 39 students taking only one of the three subjects.

Least Common Multiple, LCM

Category: Arithmetic

"Published in Newark, California, USA"

What is the Least Common Multiple (LCM) for the following numbers: 24, 36, 18, and 12?

Solution:

There are two ways in getting their LCM. I will show the both methods and let's see which method is your preference.

Method 1: You can use the intersection method. You have to list the multiples of each given numbers.

A = (24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288...)
B = (36, 72, 108, 144, 180, 216, 252, 288, 324, 360, 396.......)
C = (18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198, 216.....)
D = (12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, 192, 204, 216, 228...)

If A ∩ B ∩ C ∩ D = (72, 144, 216,.......)

Therefore, their LCM is 72. 


Method 2: You can use the continuous division method. You have to think their factors as much as you can while dividing the numbers until the quotients are all equal to 1.

          6    │   24          36           18            12
             2   │   4            6             3              2
              2   │  2            3             3              1
               3   │ 1            3             3              1
                       1            1             1              1

Therefore, the LCM is 6 x 2 x 2 x 3 = 72.

Note: Please remember how to get the LCM of the numbers very well because you will use this method later in making dissimilar fractions into similar fractions. You can add or subtract fractions if they are similar fractions where their denominators are the same. If their denominators are different or the fractions are dissimilar, then you have to make the fractions into similar fractions first by getting the Least Common Multiples (LCM) of the denominators. Once the denominators are the same, then you can add or subtract the fractions. In solving fractions, usually LCM is called Least  Common Denominator (LCD). Still, I encourage you to memorize or remember the prime and composite numbers.