Discriminant Formulas

The quadratic equation of the form $$\mid x^2 - 4x - 13 \mid = r$$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

Hence at x = 2 the quadratic equation has its minimum.

Considering the quadratic part : $$\left|x^2-4\cdot x-13\right|$$. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the function $$\left|x^2-4\cdot x-13\right|$$. Because of the modulus function, the representation of the quadratic equation becomes :

ABC'DE. 

There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.

The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :

The point C which is the value of the function at x = 2, = $$2^2-8-13$$

= -17, the reflection about the x-axis is 17.

Alternatively,

$$\mid x^2 - 4x - 13 \mid = r$$ .

This can represented in two parts :

$$x^2-4x-13\ =\ r\ if\ r\ is\ positive.$$

$$x^2-4x-13\ =\ -r\ if\ r\ is\ negative.$$

Considering the first case : $$x^2-4x-13\ =r$$

The quadraticequation becomes : $$x^2-4x-13-r\ =\ 0$$

The discriminant for this function is : $$b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$$

SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

For the second case :

$$x^2-4x-13+r\ =\ 0$$ the function inside the modulus is negaitve

The discriminant is $$16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$$

In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.

Hence $$\ 68-4r\ =\ 0$$

r = 17, for r = 17 we can have exactly 3 roots.

Given a, b, c are rational numbers.

Hence a, b, c are three numbers that can be written in the form of p/q.

Hence if one both the root is 2+$$\sqrt{\ 3}$$ and considering the other root to be x.

The sum of the roots and the product of the two roots must be rational numbers.

For this to happen the other root must be the conjugate of $$2+\sqrt{\ 3}$$ so the sum and the product of the roots are rational numbers which are represented by: $$-\frac{b}{a},\ \frac{c}{a}$$

Since the quadratic equation has negative b, it will cancel the negative sign of the sum of the roots. Hence, the sum of the roots and the products of the roots will be represented by  $$-\frac{b}{a},\ \frac{c}{a}$$

Hence the sum of the roots is 2+$$\sqrt{\ 3}+2-\sqrt{\ 3}$$ = 4.

The product of the roots is $$\left(2+\sqrt{\ 3}\right)\cdot\left(2-\sqrt{\ 3}\right)\ =\ 1$$

b/a = 4, c/a = 1.
b = 4*a, c= a.
Since b = $$c^3$$
4*a = $$a^3$$
$$a^2=\ 4.$$
a = 2 or -2.
|a| = 2

$$b^2 < 4ac$$ means that the discriminant is less than 0. Therefore, f(x)>0 for all x if the coefficient of $$x^2$$ is positive, and f(x)<0 for all x if the coefficient of $$x^2$$ is negative.

We are given that f(m)<0 and m is an integer.

So the set containing values of m will either be empty if the coefficient of $$x^2$$ is positive, or it will be a set of all integers if the coefficient of $$x^2$$ is negative.