Lesson PROOF of quadratic formula by completing the square

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<H2>PROOF of quadratic formula by completing the square</H2>

This lesson will prove that quadratic equations can be solved by "completing the square", and I will show you how it is done!

Consider the quadratic equation

{{{a*x^2+b*x+c}}} = {{{0}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

assuming coefficients &nbsp;a, &nbsp;b &nbsp;and &nbsp;c &nbsp;are real numbers.

Use &nbsp;"completing the square". &nbsp;Add and subtract &nbsp;{{{b^2/(4*a)}}}&nbsp; in the left side. &nbsp;Your equation takes the form

{{{a*x^2+b*x+c}}} + {{{(b^2/(4*a))}}} - {{{(b^2/(4*a))}}} = {{{0}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)

Regroup the terms and factor for a:

{{{a*(x^2+b/a*x+b^2/(4*a^2))}}} - {{{b^2/(4*a)}}} + {{{c}}} = {{{0}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)

The expression in parentheses is the complete square &nbsp;&nbsp;{{{(x^2+b/a*x+b^2/(4*a^2))}}} = {{{(x+b/(2*a))^2}}}.

Therefore, &nbsp;the equation &nbsp;(3)&nbsp; becomes

{{{a*(x+b/(2*a))^2}}} - {{{b^2/(4*a)}}} + {{{c}}} = {{{0}}}, &nbsp;&nbsp;&nbsp;&nbsp;or

{{{a*(x+b/(2*a))^2}}} = {{{b^2/(4*a)}}} - {{{c}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;or

{{{a*(x+b/(2*a))^2}}} = {{{(b^2-4*a*c)/(4*a)}}}.

Multiply both sides by &nbsp;{{{1/a}}}. &nbsp;You will get

{{{(x+(b/(2*a)))^2}}} = {{{(b^2-4*a*c)/(4*a^2)}}}.  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4)

An expression &nbsp;&nbsp;{{{d = b^2-4*a*c}}}&nbsp;&nbsp; is called &nbsp;<B>the discriminant</B>&nbsp; of the quadratic equation &nbsp;(1).

Assuming the discriminant is positive or zero (no negative), take the square root for both sides. &nbsp;Note we are still in the area of real numbers.

{{{x+b/(2*a)=sqrt((b^2-4*a*c)/(4*a^2))}}} &nbsp;&nbsp;or&nbsp;&nbsp; {{{x+b/(2*a)=-sqrt((b^2-4*a*c)/(4*a^2))}}}

{{{x}}} = {{{(-b)/(2*a)+sqrt(b^2-4*a*c)/(2*a)}}} &nbsp;&nbsp;&nbsp;&nbsp;or&nbsp;&nbsp; {{{x}}} = {{{(-b)/(2*a)-sqrt(b^2-4*a*c)/(2*a)}}}

{{{x}}} = {{{(-b+-sqrt(b^2-4*a*c))/(2*a)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5)

<B>Conclusion:</B>

1) &nbsp;If the discriminant of the quadratic equation &nbsp;(1)&nbsp; with real coefficients &nbsp;a, &nbsp;b &nbsp;and &nbsp;c &nbsp;is positive, 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;then the equation has two different real roots given by formula &nbsp;(5).

2) &nbsp;If the discriminant of the quadratic equation &nbsp;(1) &nbsp;is equal to zero, 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;then the equation has one real root given by formula &nbsp;(5).

3) &nbsp;If the discriminant of the quadratic equation &nbsp;(1)&nbsp; is negative, 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;then the equation has no real roots.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It follows from formula &nbsp;(4)&nbsp; because the left side as a square of the real number can not be negative.

<B>Note for the area of complex numbers.</B>

Let us consider the quadratic equation &nbsp;(1)&nbsp; in the area of complex numbers. 
This means that coefficients &nbsp;a, &nbsp;b &nbsp;and &nbsp;c &nbsp;are complex numbers and we search a solution among complex numbers. 
Then the equation always has two roots in the area of complex numbers.
These roots are given by formula &nbsp;(4).
If the discriminant is zero, &nbsp;then the two roots are actually equal each to other.

On solving quadratic equations see also other the lessons in this site

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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Learning-by-examples-on-HOW-TO-complete-the-square.lesson>HOW TO complete the square - Learning by examples</A>

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