The Theory of Quadratic Equations

A quadratic equation is a polynomial equation of second order.

A quadratic equation has two roots. The roots can also be equal and identical.

Let us write the quadratic equation in two forms AX * X + BX + C = 0 an example of a quadratic equation would be 5X*X + 3 *X + 2 = 0 Rewrite this as ( X-R1) * (X-R2) = /0.

The above step is termed as factoring.
The polynomial form of the equation is X*X + B/A * X + C/A = 0 The factored equation can be split as X * X -X( R1 + R2) + R1R2 = 0. Comparing Similar terms we can see that -(R1 + R2) = B/A R1R2 = C/A (R1 + R2) = -B/A Let us investigate the discriminant b *b - 4 * a * c b = -a ( r1 + r2) c = AR1R2; 4*A*C = 4 * A* A* R1 * R2 b*b = A*A(R1 + R2) * (R1 + R2) DISCRIMINANT = A*A(R1 + R2) * (R1 + R2) - 4*A*A*R1*R2 = A*A ( (R1+R2)((R1+R2) - 4R1R2) = A*A ( R1 - R2) * (R1 - R2).
Notice that this is a perfect square of A(R1-R2).

So if the discrimant becomes negative it means that the quadratic equation does not have real roots as squares of real numbers are also perfect squares.

Let us add A( R1-R2) to -b which is A( R1 + R2), and the sum is 2AR1. Dividing this by 2A would yield R1.
Similarly let us subtract A( R1-R2) from -b ie.
, A( R1 + R2) - A (R1-R2) which is equal to A(2R2) or 2AR2. Dividing this by 2A would yield R2. So R1 is (-B + squareroot( discriminant) ) / 2A and R2 is (-B - squareoot( discriminant) / 2A Let us take some common factoring problems that you would encounter say x * x + 5*x + 6 = 0. First step evaluate the discriminant = SQUAREROOT(25 - 24) = 1, which means that there are real roots.
The roots of the equation are (- 5 + 1)/ 2 is equal to -2 and ( -5 -1)/2 equal to -3.
The equation can be factored as (X+2)(X+3) = 0.
Let us take another example 3 * x * x + 9 * x + 6 = 0, rewriting this as x * x + 3*x + 2 = 0.

discriminant = sqrt(9-8) = 1 R1 = -1 and R2 is -2.
So the factored form of the same equation is (x + 1)( x+ 2) = 0. A quadratic equation can also be plotted on a graph. It will yield the equation of a parabola.