Quadratic Equation
Introduction To Quadratic Equations: Topics Included
A quadratic equation is something fun and easy to learn and master. solving a quadratic equation is almost quiet robotic, because we can follow a definite algorithm with very slight, but visible alterations to solve different types of equations. In this discussion about the quadratic equations we will get to know about
 The general form of a quadratic equation
 The general solution to a quadratic equation
 Solving a quadratic equation by factorization
 Solving a quadratic equation by completing the square
 Real and imaginary roots
 The relationship between roots and coefficients
The discussion is introduced with a beginner to quadratic equations in mind. Therefore you can get the basic knowledge in a confident way by following the discussion below. However you will find the discussion useful even if you have already learnt quadratic equations, because the representation here might help you to clarify more about certain areas.
The general form of a quadratic equation
Quadratic equations, as the name implies, mean equations of second degree. We can define the quadratic equation quiet mathematically considering it as a function of a certain variable x, which has two solutions for x, real or imaginary, but not both, when that particular function equals to the zero. However we are not going to use much of the mathematical jargon, because our aim in this discussion is to understand the concepts easily and fully, without confusing with a set of mathematical words.
So a simple quadratic equation is a kind of equation with a variable of a second degree. (Something like x^{2} + as the highest degree of the equation). This can be written down more generally as
ax^{2 }+ bx + c = 0; where a, b and c are real numbers.
Solving the equation simply means that we find the values for which the left hand side (L.H.S) equals to the zero.
For example take x^{2}  6x +8 = 0
To solve this equation we are required to find the values for which L.H.S equals zero. We call those values, the solution to this equation. To clarify this more, let’s substitute 2 to the x in the L.H.S of the above equation.
Then
(2)^{2 }– 6 (2) + 8 = 4 – 12 + 8
= 0
Therefore we see 2 is a solution to the above equation. We can call it a root of the above quadratic equation. Similarly you would find 4 too becomes a root of the above equation. And that’s all. Remember every quadratic equation has only 2 roots at the most! So how did we find these two roots here? We will discuss about it few minutes later.
Now just remember the general form of the quadratic equation is ax^{2 }+ bx + c = 0, where a, b and c are real numbers. That’s what we would find everywhere in this discussion and in other discussions about quadratic equations as well. The word ‘general form’ stresses that it is used with a universal context.
The general solution to quadratic equation.
Here the letters a, b and c are the coefficients used earlier in the general form of the quadratic equation.
When we substitute correctly the numeric values to a, b and c we can get the numerical solution of the particular equation we are considering. For instance if we consider the above example
We have a= 1, b= () 6 and c= 8. It’s a very simple context, not a big deal. If somebody asks you to solve a particular equation all you have to do is to substitute the values accordingly and you will come up with the answer! But, wait a minute, because there are many easier methods to solve a quadratic equation than just substituting!
We will consider them in the following sub topics in our discussion, and also we can discuss some harder problems where you can not directly apply the above formulae to come up with an easy answer! Do not worry though because we also learn how to solve those difficult equations as well, from our discussion!
Solving a quadratic equation by factoring
This is a very convenient method to solve quadratic equations both simple and hard quadratic equations. It includes nothing but the basic factorization you learn in high school. Probably you have heard of factorization of prime numbers etc. etc. The only difference here is that instead of prime numbers we come across linear functions (Do not worry over the word, it only means something in the form of px+q, where p and q are real numbers).
So for example if we consider our previous equation x^{2}  6x +8 = 0
We can rearrange the L.H.S of this particular equation as follows.
x^{2}  6x +8 = (x2) (x4)
You can expand the right hand side and check if I am correct! Now how did we factor this is another matter and we would discuss it shortly. For now, just look at the basic modified shape of the previous quadratic equation. We altered the form of the equation so that it becomes a product of two simple linear functions, in this case (x2) and (x4). Now this slight change of the shape helps us a lot in solving our equation.
For,
x^{2}  6x +8 = 0
(x2)(x4) = 0
Therefore either x2 =0 or x4 = 0, because it is very clear to you that we will get 0 in the other side by the product of two numbers if and only of one or both of these numbers are 0. Simple mathematics, right?
So we can say x = 2 and x= 4 satisfy the above quadratic equation, and are therefore the roots of the above quadratic equation.
Now let’s focus on to factoring quadratic equations (a quadratic expression, to be precise)
How to factor quadratic equations?
Now we introduce you another word related to quadratic equations, ‘Discriminant’. Don’t worry too much over it because it’s just a name!
In fact discriminant = b^{2 }– 4ac
Simple, isn’t it? This is what you found under the square root part of out general solution to a quadratic equation.
We can denote this discriminant by the Greek letter capital Delta (A small triangle). In the foregoing parts of our discussion I will use the letter capital D to denote Discriminant (It makes me easier to type this document but you will find the symbol Delta in many other references, and do not confuse)
Now we will understand factoring quadratic equations
1. Look at the discriminant (You do not see it at once, just calculate it mentally)
2. If it is a square of a rational number the quadratic equation has two real and distinct rational roots
3. Think about the possibilities to make the number c (in this example +8) by the product of two whole numbers. You will come up solutions, like 8*1, 2*4.
The next condition is to find these numbers such that the sum of these two numbers equals the minus value of existing coefficient of x (In this example 6 because minus value of – 6 is +6)
4. You have just found the factors, and the roots of the quadratic equation! Wow!
5. Note: If D is not a square of a rational number still you come up with solutions but the roots are no longer rational.
It’s not a big deal. Let’s have a look.
We write down the possibilities for making +8
1*8
2*4
2 * 4
½ * 16 and so on. The listing is seemingly endless.
.
Now we think about our second condition. The numbers are found such that their sum equals the coefficient of x.
So we dig deep again. No there is nothing hidden that deep down for us to discover. We find the answer almost instantly. Now of all the numbers above there is only one pair of numbers that give 6 upon adding, and they are only 2 and 4. So our solutions are 2 and 4 and our factors become x2 and x4. Not something difficult.
So you can utilize this simple method to factorize the quadratic expression to the left and therefore solve the quadratic equation. This method is particularly useful when the D is a perfect square, over substituting directly to the general solution.
Solving quadratic equation by completing the square
For this method you don’t have to worry whether D is a square of a rational number or not. In fact this is the method used to derive the general solution to the quadratic equation, which we discussed above. This is not a strange or difficult thing to understand. It just involves using the method called difference of two squares to facto the quadratic expression, and that’s all, really.
Let’s understand this using an example.
Remember (p+q)^{ 2} = p^{2} + q^{2} + 2pq
Now consider x^{2}  6x +8=0 (This is the same example we used earlier. Of course you can use any example for this)
Now follow the procedure below
1. Divide the coefficient of term x by two
2. Write (x + the quotient) with in a bracket and square it.
3. Expand it mentally using the above mentioned rule
4. Add or subtract enough amounts from the constant term to match the previous constant term
5. Use difference of two squares
Let’s see how this is done. It’s actually very easy
1. I divide the coefficient of x by two.
() 6 / 2 = () 3
2. I write (x+ quotient) within brackets and square it. I get
(x3)^{2}
3. I expand it mentally. I can see I get x^{2}  6x + 9.
4. Now I subtract 1 from it to match the constant term of the previous expression. That is to get 8, because we are solving x^{2}  6x +8=0
Therefore I can write x^{2}  6x +8= (x3)^{2} – 1
5. Now I use the difference of two squares. That is a^{2} – b^{2 }= (a+b) (ab)
Therefore I get (x3 + 1) (x 3 – 1) = 0
Or (x2) (x4) = 0
So we see we have come up with the roots of the equation again.
Real and imaginary roots
If D is zero or a positive amount we call the equation has real roots. If D is negative we call the equation has imaginary roots.

Imaginary roots are written using complex numbers and are not to be confused here. What you should know is that the quadratic equation can have two roots at the most. They can be either real, or complex (imaginary). But they can’t be both. That is one root can not be real while other is complex. That never happens.

You should also know that an imaginary roots take the general form of (a + i b) where a and be are both real numbers and i stands for iota.

Another interesting thing to know will be that complex roots are conjugate. That is if ( a + i b ) s a root of a certain quadratic equation then ( a i b) is also a root of that equation, In fact then the roots of that particular equation are ( a + i b ) and ( a  i b) because we already know that quadratic equations have two roots at the most.

Now there is a special case when D = 0. Then we only get one real root as the solution for the quadratic equation and then we call the roots are real and equal (in other cases we call roots are real and distinct or just distinct as the occasion demands)
For example consider the equation x^{2} + 6x + 9 = 0
Clearly we see D= 0 here.
We can follow our previous procedure to factorize the expression in the L.H.S.
You will get ( x + 3 ) ^{2} = 0 (Try to do this. It’s so simple)
Now clearly ( x+3) = 0 and therefore x can have one and only one real value and it is – 3 here.
The relationship between roots and the coefficients
Roots and coefficients of a quadratic equation hold a good relationship. Now let’s take p and q as the roots of a particular quadratic equation.
Then
Coefficient of x / coefficient of x^{2} =  (p + q); Note the minus mark
And Constant term /coefficient of x^{2 }= pq
This relationship between roots and coefficients is very useful to solve many problems related to quadratic equations.
So we are at the end of our brief discussion about quadratic equations. Here we learnt about the general solution, factoring and completing the square to solve a quadratic equation and also about real and imaginary roots and at last about the relationship between roots and coefficients of a quadratic equation. I hope you enjoyed our lesson. If you liked this you can bookmark this and you can share this with your friends on facebook and google+. Let me know about our lesson by voting below. Thank you :)
Comments
I tried really hard to follow you, I really did. I have a PhD in history, but I barely managed to get a C in college Algebra (and that was 30 years ago). Is there anyway to make it even simpler, or maybe divide this Hub into two or three sections? Just asking.
Have a wonderful Christmas.