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## The Theory of Quadratic Equations

A quadratic equation is a polynomial equation of the second order. A quadratic equation has two roots. Roots can also be equal and even. Let’s write the quadratic equation in two ways

AX * X + BX + C = 0 an example of a quadratic equation would be 5X*X + 3 *X + 2 = 0

Let’s rewrite the quadratic equation as (X-R1) * (X-R2) = 0. The above step is called factoring.

Let’s rewrite the original quadratic equation as X*X + B/A * X + C/A = 0

The equation can be rewritten as X * X – X( R1 + R2) + R1R2 = 0.

Comparing like terms we can see that -(R1 + R2) = B/A

R1R2 = C/A

(R1 + R2) = -B/A

Let’s investigate 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)

DIVISION = A*A(R1 + R2) * (R1 + R2) – 4*A*A*R1*R2

= A*A ( (R1+R2)((R1+R2) – 4R1R2)

= A*A (R1 – R2) * (R1 – R2).

Note that this is a perfect square of A(R1-R2). So if the discalment is negative it means that the quadratic equation has no real roots as the squares of real numbers are also perfect squares.

Let’s add A(R1-R2) to B which is A(R1 + R2), and the sum is 2AR1. Dividing this by 2A will yield R1.

Similarly let’s take A(R1-R2) from B i.e., A(R1 + R2) – A (R1-R2)

which is equal to A(2R2) or 2AR2. Dividing this by 2A will yield R2.

So R1 is (-B + squareroot( calucal) / 2A and R2 is (-B – squareroot( calulo) / 2A

Let’s take a look at some common problems you may encounter

say x * x + 5*x + 6 = 0.

The first step tested the discriminant equal to SQUAREROOT(25 – 24) = 1, which means there are true roots.

The roots of the equation are (- 5 + 1)/ 2 equals -2 and (-5 -1)/2 equals -3.

The equation can be expressed as (X+2)(X+3) = 0.

Let’s take another example

3 * x * x + 9 * x + 6 = 0, rewriting this as x * x + 3*x + 2 = 0.

quotient = sqrt(9-8) = 1

R1 = -1 and R2 is 2. So the integrated form of the same equation

(x + 1)(x+ 2) = 0.

A quadratic equation can also be plotted on a graph. When plotted it will produce the equation of a parabola.

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