## Meaning of Quadratic Equation and its basic proof

If you have landed on this page for the first time, know you are on the right direction of gaining full understanding towards quadratic formula development. Many teachers of algebra have had presentations all of which converge into similar approach. But according to my research I have done over decades, a big number of materials you find online fall short of unleashing the back-end concept of completing the square in finding roots of quadratic equations. Well said, lets’s get started.

## The general form of quadratic equation is given by

$\fn_jvn&space;\large&space;{\color{Red}&space;\mathbf{ax^{2}+bx+c=0}}$

Where a≠0 and if it is equal to zero, then the quadratic equation does not exist; most importantly a, b and c are constants. Some scholars proudly call them numerical coefficients. In this case x is a variable and c is the constant term. At this stage I hope you have understood the principles of quadratic equations.

## Basic method of completing the square

Completing the square method is a method of solving quadratic equations, where polynomials of second degree are expressed as a product of linear factors . For quadratic equations that can not be factorized, completing the square method is technically essential. Then, zero-product property is applied usually equating product of the linear factors to zero.

## What you need to know before completing square by standard method

• One algebraic identity
• Zero-product property

It should be noted that in this method, we are going to use only one identity.

$\fn_jvn&space;\large&space;\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{a^{2}+2ab+b^{2}=\left&space;(&space;a+b&space;\right&space;)^{2}}}$

The zero-product property asserts that the product of two nonzero elements is nonzero. That is to say, if (a+b)(a-b)=0, then (a+b)=0 or (a-b)=0.

The standard method of completing the square requires that you divide by a through the equation.

$\fn_jvn&space;\large&space;\mathbf{ax^{2}+bx+c=0}$

By doing so produces

$\fn_jvn&space;\large&space;\mathbf{x^{2}+\frac{b}{a}x+\frac{c}{a}=0}$

The critical question now arises. Why divide by a through the equation? It is to simply and concisely make the general quadratic equation mimic the order of algebraic identity which is on one hand a complete square and the other a perfect sqaure. In nuthshell, a is divided through the equation so that the first of the quadratic trimonial is a perfect square; x squared.

The algebraic identity in play here is shown below. There are three important algebraic identities that will be covered later in advanced methods of proofing quadratic equations.

$\fn_jvn&space;\large&space;{\color{Magenta}&space;\mathbf{a^{2}+2ab+b^{2}=\left&space;(&space;a+b&space;\right&space;)^{2}}$

The identity establishes that a is a variable and b is constant, b squared is a parameter as well . After careful observation, the coefficient of a squared is 1. Therefore, a is divided through the general equation of quadratic form to make the co-efficient of x also equal to 1. As for my research making the co-efficient of the first term equal to 1 in a polynomial of second degree is a weak axiom. In my next approach of completing the square, it  is demonstrated that a is divided through the general equation so as to make the first term a perfect square. In my opinion this is the strongest axiom in unearthing the concepts of completing the square of polynomials of second degree.

Now, let’s put more focus on the first two terms of the equation as they comprise the variable . The first two terms are crucial in completing the square on the LHS.

$\fn_jvn&space;\large&space;\large&space;{\color{Black}&space;\mathbf{x^{2}+\frac{b}{a}x}}{\color{DarkRed}&space;\mathbf{+\frac{c}{a}=0}}$

On reference to the algebraic identity , what must be added to the first two terms to complete a square? In order to make a perfect square trinomial you need to calculate half the coefficient of x, square the result and add to both sides of the equation.

Another important question emerges immediately. Why half the coefficient of x? From the algebraic identity, the middle term is 2ab and by associative property of multiplication if you rewrite in standard form, 2ab=(2b)a where 2b is the co-efficient and a is the unknown variable. Given the first two terms of the identity and you suppose to complete the square you have to take half the coefficient of a which is 2b . Take half of 2b to get rid of 2 in the co-efficient of. In quadratic equations, particularly trinomials, half the co-efficient of the second term was invented to get rid of inherent 2, squaring the result produces the third term in a process of completing the square .

Let’s quickily start the process of completing the square. Take half the coefficient of x , square the result and add to both sides of the general quadratic equation.

The left side of the equation below is now a complete square

$\fn_jvn&space;\large&space;\mathbf{x^{2}+\frac{b}{a}x+\left&space;(&space;\frac{1}{2}\cdot&space;\frac{b}{a}&space;\right&space;)^{2}=-\frac{c}{a}+\left&space;(&space;\frac{1}{2}\cdot&space;\frac{b}{a}&space;\right&space;)^{2}}$

Since you have completed the square on the LHS, express it as a perfect square on the same side in the next step

## Full understanding of complete square and perfect square

$\fn_jvn&space;\large&space;\\\textbf{complete&space;the&space;square&space;on&space;the&space;LHS&space;in&space;the&space;form}\\\\&space;{\color{Magenta}&space;\mathbf{a^{2}+2ab+b^{2}}}\\\\&space;\textbf{express&space;as&space;a&space;perfect&space;&space;&space;square&space;as&space;shown&space;below}\\\\&space;{\color{Magenta}&space;\mathbf{\left&space;(&space;a+b&space;\right&space;)^{2}}}\\\\&space;\textbf{combine&space;the&space;two&space;to&space;create&space;algebraic&space;identity}\\\\&space;{\color{Magenta}&space;\mathbf{a^{2}+2ab+b^{2}=\left&space;(&space;a+b&space;\right&space;)^{2}}}\\$ $\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;x+\frac{b}{2a}&space;\right&space;)^{2}=-\frac{c}{a}+\left&space;(&space;\frac{b}{2a}&space;\right&space;)^{2}}$

As you can see, the left side above is now a perfect square.

Similarly, all elements or terms on the right side of the equation are constants. Therefore, it is much easy and straight forward to extract solutions from them by taking square roots-you will find more details below. And the left side comprise of unknown parameter(s).

On the RHS find the LCM and simplify the fractions by addition or subtraction.In this case the LCM is

$\fn_jvn&space;\large&space;\mathbf{4a^{2}}$

Which gives a simplified fraction as shown on the RHS

$\fn_jvn&space;\large&space;\mathbf{\left&space;(&space;x+\frac{b}{2a}&space;\right&space;)^{2}=\frac{&space;b^{2}&space;-4ac}{4a^{2}}}$

Next step, take square roots on both sides of the equation to solve the linear form inside the brackets on the LHS with unknown variable.

$\fn_jvn&space;\large&space;\mathbf{\sqrt{\left&space;(&space;x+\frac{b}{2a}&space;\right&space;)^{2}}=\pm&space;\sqrt{\frac{b^{2}-4ac}{4a^{2}}}}$

You have noticed that taking the square roots results two possible solutions , plus(+) or minus() combined in one as ±.

$\fn_jvn&space;\large&space;\mathbf{x+\frac{b}{2a}=\pm&space;\sqrt{\mathbf{\frac{b^{2}-4ac}{4a^{2}}}}}$

The equation above is reduced and simplified to linear form, now find x by substracting the constant part on the left from both sides.

$\fn_jvn&space;\large&space;\mathbf{x+\frac{b}{2a}-\frac{b}{2a}=\pm&space;\sqrt{\mathbf{\frac{b^{2}-4ac}{4a^{2}}}}-\frac{b}{2a}}$

Now, find LCM and simplify the RHS

$\fn_jvn&space;\large&space;\mathbf{x=\frac{-b\pm&space;\sqrt{b^{2}-4ac}}{2a}}$

That is it !