Solution of a quadratic equation by completing the square

One method of obtaining the roots of a quadratic equation is the factoring method. Another method that is used is completing the square method.

Consider the following situation:
The product of Sara’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?
To answer this, let her present age (in years) be = x. Then the product of her ages two years ago and four years from now is (x-2)*(x+4).
Therefore, (x-2)*(x+4) = 2x+1
I.e, x^2 + 2x – 8 = 2x + 1
i.e, x^2 – 9 = 0

So Sara’s present age satisfies the quadratic equation x^2 – 9 = 0
We can write this as x^2 = 9. Taking square roots, we get x = 3 or x = -3. Since age is always a positive number, x = 3 would be our answer.

Now consider the quadratic equation (x+2)^2 – 9 = 0. To solve this we can write it as (x+2)^2 = 9. Taking roots we have x+2 = 3 and x+2 = -3. Therefore x = 1 or x = -5. These are the roots.

In both the examples above, the term containing x is completely inside a square, and we found the roots easily by taking square roots. But what if we are given the equation x^2 + 4x – 5 = 0 to solve? We would probably apply factorization unless we (somehow!) realize that x^+4x-5 = (x+2)^2 – 9.

So solving x^2+4x-5 = 0 is equivalent to solving (x+2)^2-9 = 0, which we have seen is faster. In fact, we can convert any quadratic equation to the form (x+a)^2 – b^2 = 0 and then we can easily find its roots. This process is called completing a square.

Suppose we were to complete the square for x^2+4x. The process is as follows:
X^2+4x
=(x^2 + (4/2)x) + (4/2)x
= x^2+2x+2x
= (x+2)x + 2*x
= (x+2)x + 2*x + 2*2 – 2*2
= (x+2)x + (x+2)2 – 2*2
= (x+2)(x+2) – 2^2
= (x+2)^2 – 4
So, x^2+4x-5 = (x+2)^2 -4 -5 = (x+2)^2 – 9.
So turning x^2+4x-5 to (x+2)^2-9 makes a complete square of the expression.

In brief, this can be shown as follows:
X^2+4x = (x+(4/2))^2 – (4/2)^2 = (x+(4/2))^2 – 4
So, x^2+4x-5 = 0 can be rewritten as
(x+(4/2))^2 -4-5 = 0
i.e., (x+2)^2 – 9 = 0. This was the completing square method in brief.

Addition of like fractions

We know that whole numbers can be added, subtracted, divided or multiplied. Similarly fractions can also be added, subtracted, divided or multiplied. A problem of addition of fractions can be of either of the two types:
(a) Adding fractions like denominators
(b) Adding fractions – unlike denominators

Adding like denominators fractions:

Adding fractions with like denominators word problems may be worded using different terms such as: “How to add fractions with common denominators?’ or ‘adding fractions with same denominators’.  These words, “common denominators” or “ same denominators” mean the same thing as “like denominators”. By definition, if two fractions have the same or equal denominators they are called, like fractions.

For being able to add fractions, the denominators have to be equal or same. In other words, for adding fractions it is important that all the addends are like fractions and have the same common denominator. For adding fractions with unlike denominators, we first make the denominators equal by taking the lowest common multiple of the denominators and converting each of the fractions to its corresponding like fraction that has denominator equal to the lowest common multiple found.

For adding like fraction, the following steps are to be followed:

Step 1: Make sure that the fractions are like fractions. That means, ensure that the denominators of all the fractions to be added are equal.
Step 2: Add up all the numbers on the numerators of all the fractions.
Step 3: Put the result found in step 2 over the common denominator that we confirmed in step 1.
Step 4: Simplify the fraction if needed. That is, reduce the fraction to its lowest terms if needed.

Let us illustrate the above process with an example:

Example 1: Add 1/10, 3/10, 1/10

Solution:
Step 1: The fractions to be added are 1/10, 3/10 and 1/10. Note that the denominators of all the three fractions are same and equal to 10. So we can move on to step 2.

Step 2: The numerators of the fractions are 1,3 and 1. Adding them we have,
1 + 3 + 1 = 5.

Step 3: The result found in step 2 was the number 5. We put this number over the common denominator that we confirmed in step 1 which was 10. So we have,
= 5/10

Step 4: Now we reduce the fraction thus found to its lowest terms. So,
5/10 = (5*1)/(5*2), the common factor 5 cancels out and we are left with,
= ½ <- answer.="answer." p="p">

Word Problems on Addition

Word problems in mathematics mean a text representation of a mathematical expression. Word problems make it easy for one to solve a problem. Words problems include all the four mathematical operations – addition, subtraction, multiplication and division. Let’s have a look at word problems on addition in this post.
Word problems on addition are nothing but a text representation of addition operation. For example: Maria bought nine bed rails from Bed Guard India

collection online. Shiva gave three Bed Guard India bed rails to Maria. How many number of bed rails is Maria left with? This problem is the word problem of addition operation (9+3=?)

Solving word problems on addition:

Solving word problems on addition include three steps. Firstly, identify the numbers to be added. Secondly, convert the word problem into a mathematical additional operation and finally add the numbers. For example: Patrick bought two Angry Bird toys from children online shopping store. Joe bought five Chota Bheem toys from the same children online shopping
store. He gave these five toys to Patrick. How many toys do Patrick has now? Identifying the numbers, Patrick bought 2 toys and Joe bought 5 toys. Converting the word problem into mathematical operation, (2+5); adding the numbers (2+5=7). Thus, the answer is 7 toys.

Examples of Word Problems on Addition

1. Mary has one baby bottle from Bottle Sterilizer India collection. Joe bought 2 baby bottles from Bottle Sterilizer India
from online store and Thomas bought 5 baby sterilizers from local store. Three of them gave these to Philip. How many baby bottles and sterilizer are there in total with Philip?Answer: 1 baby bottle + 2 baby bottles + 5 baby sterilizers
(1+2+5) = 8 baby products from Bottle Sterilizer India
collection.2. There are five play schools in the city. One play school is in the process of construction and Thomas is planning to complete 2 play schools in two different corners of the city by the end of next year. How many play schools will be there by the end of next year?
Answer: 5 play schools + 1 constructing play school + 2 planned play school
(5+1+2) = 8 play schools in the city by the end of next year.
These are examples of word problems on addition.