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I need to understand the proof of the quadratic equation. I know that it can be done using a completing the square method. Is it possible to do this quadratic equation proof without completing the square?

ax 2 + bx + c = 0

where a, b and c are constants with a not equal to zero.

...Solve the above equation to find the quadratic fomulas

Given

ax 2 + bx + c = 0

Divide all terms by a

x 2 + (b / a) x + c / a = 0

Subtract c / a from both sides

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

and simplify

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

Add (b / 2a) 2 to both sides

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

to complete the square

[ x + (b / 2a) ] 2 = - c / a + (b / 2a) 2

Group the two terms on the right side of the equation

[ x + (b / 2a) ] 2 = [ b 2 - 4a c ] / ( 4a2 )

Solve by taking the square root

x + (b / 2a) = ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) }

Solve for x to obtain two solutions

x = - b / 2a ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) }

The term sqrt { [ b 2 - 4a c ] / ( 4a2 ) } may be written

sqrt { [ b 2 - 4a c ] / ( 4a2 ) } = sqrt(b 2 - 4a c) / 2 | a |

Since 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written

x = [ -b + sqrt( b 2 - 4a c ) ] / 2 a

x = [ -b - sqrt( b 2 - 4a c ) ] / 2 a

The term b 2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible

case 1: If b 2 - 4a c > 0 , the equation has 2 solutions.

case 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2.

case 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions.

...

Your -b/a value is the sum of your roots