Quadratic equation, the Vedic way

Solving quadratic equation by a special technique-Symmetric

This wonderful method was stumbled upon by me while I was looking for a way to better explain quadratic equation. I discovered the traditional way of solving quadratic equation might not be enough for some people. Better still a more convenient method may be appealing to some people.
So in my search I came across the famous Vedic math’s India where I learn the technique I will be sharing with you.
Hope you haven’t forgotten that a quadratic equation is a polynomial of the second degree. Sometimes you will be given a simple equation that will lead to a quadratic equation.

For example:

X + 1/X = 26/5

X^2 + 1 =26X/5 which implies that 5X^2 +5 = 26X

Bringing 26x to the L.H.S of the equation we have

5X^2 – 26X + 5 = 0 (A clear quadratic equation)

From here now we will want to apply any of the methods we have learn so far to solve this quadratic equation.
Let us apply the formula method which we consider a typical or prototype solution of quadratic equation.

Recall: X=(-b+or-sqrt(b^2-4ac))/2a

Now setting a=5 , b=26 ,c=5
So that:

X=-(26 + (26^2 – (4*5*5))/2 or X=-(26 - (26^2 – (4*5*5))/2

X=5 or X=1/5

Oh yes that’s the solution to that quadratic equation but wait a minute .what if you can avoid all the trouble of going through those steps. What if you can tell the solution by merely looking at the equation. I bet it’s possible and that’s just what I learn on Vedic maths .I am going to teach you exactly what I learn and the way I understand it.

Now let’s take our former example

X + 1/X = 26/5

26/5 is the same as 5 + 1/5 (come on I just divide 26 by 5 which gives you 5 remainder 1 )

X + 1/5 =5 + 1/5

By looking at the symmetric nature of the equation now, we can easily say that

X=5 or X=1/5

Very simple indeed compare to the trouble we go through when we solve the same equation by formula method that we applied to the equation in the first place
Let’s do more of it to better open your understanding.

X + 1/X = 10/3

Oh yes 10/3 is the same as 3+1/3

X + 1/x = 3 + 1/3
By looking at the symmetric nature of the equation now,you can instantly tell the answer.

X=3 or X=1/3 so simple,isn’t it?

Well, I know you may be asking yourself, what if the equation didn’t come in this simple form as the one’s explained above.
Your question is not out of place in fact I am going to show you how to deal with any quadratic equation right away

Take for an example: 5X^2 – 26X + 5 = 0

What you do is divide through first by coefficient of X raise to the power of 2 which is 5 in this case

The equation becomes: X^2 – 26X/5 + 1 = 0

Secondly divide through by X. The equation now becomes:

X – 26/5 + 1/x = 0

Now move -26/5 to R.H.S

X + 1/X = 26/5 (always be mindful of the sign, when positive travel to the other side it turns – and vice versa)

Now you can take it from there. But let’s finish it up together

X + 1/X = 5 + 1/5

Obviously, from the symmetric nature of the equation now,

X=5 or X=1/5

Simple as ABC. What you do when you see a quadratic equation is to try to reduced it to a simple form as we do above so you can solve it quickly.
Looking at symmetric nature of the equation is the technique applied here. It is not limited to just simple equation as above. Let’s consider one more example and we will round up.

(3x + 2) + 1/(3x + 2) = 65/8

Solution:

(3x+2 )+ 1/(3x+2) = 8 + 1/8

You can see from the symmetric nature of the equation that

(3x +2)= 8 and (3x +2) =1/8

I know you can solve that simple equation

X= 2 or X=17/24 that’s what you get when you solve for x

Okay we haven’t take an equation with a negative sign since. Let’s take one

(3x + 2) - 1/(3x + 2) = 63/8
Very similar to the one above but be careful and observe that 63/8 is same as 8-1/8 that’s the trick there. So that:

(3x + 2) - 1/(3x + 2) = 8 - 1/8

Obviously from the symmetric nature of the equation.

(3x +2) = 8 and (3x +2) = -1/8

X= 2 or X= -17

Hope you enjoyed the lesson. Thanks to Vedic math's

Summary:

Reduce the equation to it’s simplest form

Look at the symmetric nature’s of the equation and solve for x

That’s all that is to the technique we learn this time

Exercise: Try the following; leave your answer in the comment box

1. 2x/(5x+1) – (5x +1)/2x = -15/4

2. (4x + 3)/(3x+4) – (3x + 4)/(4x + 3) = 24/5

Answer will be posted after reactions.

The Completing the square method - Quadratic Equation

Completing The square method of Quadratic Equation
Another quick way of solving quadratic equation as earlier stated is by using completing the square method .
It is so simple if you had understand how we derived our formular method of the quadratic equation.
Remember quadratic Equations are polynomial of the second degree other and is form can be represented as below:
Ax^2 + Bx+ C =0
Some quadratics are very simple to solve because they come in simple form like below:
Say (x-3)^2=9
This type of quadratic equation could quickly be solved by taking square root of both sides of the equation.
i.e sqrt ( x-3)^2 =sqrt(9)
x-3=+0r-3 (note that when you take a square root of a number say 9 for example,the result would be either + 0r - )
By solving for x in the equation above we going to have two answers.
i.e x=3+3 or x=3-3
x=6 or x=0
But,what about the situation when our equation do not come in this form. Most quadratic equation will not come neatly squared like this. In this case you first use your mathematical technique to arranged the quadratics in a neatly squared part equals to a number like the example treated above. Thus, the completing the square method.

For a typical Example:
Solve the quadratic equation 4x^2 -2x-5=0
Solution
Step 1: move -5 to the R.H.S to the equation (R.H.S-right hand side)
4x^2-2x=5 (remember when you move -5 to the other side of the equation it becomes +5)
Step 2: Divide through by the co efficient of your X squared term (which is 4 in our example)
The equation now becomes:
X^2 – ½X = 5/4
Step 3: T ake half the coefficient of X term and square it and add it to both sides
½ of -1/2 =-1/4
When you square it you have 1/16 add to both side of the equation which now becomes:
X^2 - 1/2X + 1/16 = 5/4 + 1/16
Step 4: Convert the left hand side to a squared form and simplify the R.H.S
(x-1/2)^2 = 21/16 (now you have a simple squared form just like our first example)
Step 5: Find the square root of both sides
x-1/2 = + or – sqrt(21/16)
solving for x finally leads to 2 answers either :
X=1/2- sqrt(21/16) or X= ½ + sqrt(21/16)


Congratulations you have successfully complete the steps for solving a quadratic equation using completing the square method.
Summary:
Move the number part to the right hand side of the equation
Divide through by the coefficient of the x squared term
Take one-half the coefficient of the x term ,square it and add it to both sides of the equation
Re-arrange your equation by putting the right hand side in squared form and simplifying the left hand side
Take the square root of both sides remembering the + or – sign on the right hand side
Finally solve for two possible values of X
Exercise:
Solve X^2 +6X-7=0 by completing the square method
Post your answer in the comment box
You can also subscribe to this post to see solutions to the above questions by people

Graphing quadratic equations/functions

Another intresting way of solving quadratic equation is discuss in below it is called the graphing method
The Math Blog: Graphing quadratic equations/functions

The Formula Method of The Quadrtic Equation - How It Was Derived

The formular method of the quadratic equation is often a popular request by student. So we have taken our time to break down the steps to how this formular was arrived at.

How the formular method was derived.

The formular method of the quadraric equation can be regarded as a derivative
Of completing the square method of solving the prototype of a quadratic equation.
I.e ax^2 + bx + c = 0

By solving the prototype quadratic equation above ,using completing the square method. We will arrive at the quadratic formular below

(-b +-_ /b^2-4ac)
2a
Now let's examine how the formular came about.

Take your equation (the quadratic equation prototype)

ax^2 + bx + c = 0

move the constant c in the quadratic equation to the R.H.S of the equation (R.H.S - right hand side)

we have :
ax^2 + bx = -c

now divide through by a

ax^2/a + bx/a = -c/a

we will arrive at :

x^2 + bx/a = -c/a

applying Completing the square method

x^2 + bx/a + (b/2a)^2 = - c/a + (b/2a)^2 ---------(1)

from the above equation,it could be pictured that half the co-efficient of x (of bx/a) which is b/2a was squared and addeto to both sides of the equation.(That is what we call completing the sqare method of the quadratic equation)

by solving equation (1):

we will have,

(x+b/2a)^2 = b^2/4a^2 - c/a

{note that (x+b/2a)^2 when expanded is the same as x^2 + bx/a +(b/2a)^2 }

Acting on the R.H.S of the equation,

(x+b/2a)^2 = b^2 - 4ac
4a^2

taking the square root of both sides we have,

x + b/2a = +-/b^2 - 4ac
2a


now solving for x we have,

X= -b/2a +- /b^2 - 4ac
2a
which could further be simplified as



X= -b+-/b^2-4ac
2a


i.e. X= -b+/b^2-4ac or X= -b-/b^2-4ac
2a 2a


That's just how the formula method of quadrtic equation was derived.

The Quadratic Equation - Factoring Method

The quadratic equations is a topic under the series of algebra. quadratic equation is a polynomial of second degree.

quadratic equation is a popular topic in mathematics.

Quadratic equation is of the form ax*2+bx+c =0

Where a is the co-efficient of x raise to the power of 2
in the quadratic equation

b is the co-efficient of x in the quadratic equation

While c is a constant in the quadratic equation.

An example of a quadratic equation is:

X*2 + 3X + 2.=0

The quadratic equations can be solved by one of the following methods:

1. Factoring method
of solving quadratic equation

2. Completing the square method
of solving quadratic equation

3. Graphical method of solving quadratic equation

4. Formular method of solving quadratic equation

Factoring method of solving quadratic equation is more of elementary methods of solving quadratic equations and it is easy to use .
It involves taking factors of the quadratic equation by multiplication of cofiecient of X(square i.e ,a) in the quadratic equation and C (which is a constant) in the quadratic equation and try to bring out a combination such that when added or subtracted will result in co-efficient of X( raise to the power of 1)in the quadratic equation as in AX*2 + BX + C =0
which is the prototype of our quadratic equations

Let us take a typical example of the quadratic equation we have been discussing above

Question: Solve by factorisation method the quadratic equation X*2 + 3X + 2=0

Solution:
compare the quadratic equation X*2 + 3X + 2=0 to it's prototype quadratic equations AX*2 + BX + C =0
Where A=+1
in our quadratic equations example
B=+3
in our quadratic equation example
C=+2 in our quadratic equation example
As in the coefficients of our quadratic equation prototype when compared.

Now multiply +2 and +1(obtained from our quadratic equation example) (multiplying A and C)
The result is +2
Factors of +2 include +2 and +1 it could also be -2 and -1 because when multiply wa still arrive at +2

now that we have our factors for our quadratic equation,

we will re-write the quadratic equation as

x*2 + 2X + X + 3=0

Note that 2X + X = 3X

So that no mathematical rule is violated or value changed in our quadratic equation.

The quadratic equation x*2 + (2X + X) + 3 Is still the same as quadratic equation x*2 + 3X + 3

If we decide to group our quadratic equation x*2 + 2X + X + 3=0

we would then have:
(X*2 + 2X)+(X+2)=0
X(X+2) + 1(X+2)=0
(X+1)(X+2)=0

i.e. Either (X+2) or (X+1) = 0
X+2 =0, or X+1=0.
X=-2 or X=-1

Factorisation method of solving quadratic equation has previously said, is more of elementary method in solving quadratic. Some other quadratic equations may not moveable by factoring method of quadratic equation so we can employ other methods of solving quadratic equation like the

Completing the square method of solving quadratic equation,graphical method quadrtic equation, or formula method of solving quadratic equation.