Conditions For Specific Roots Of A Quadratic Equation

Theorem: For a quadratic equation $ax^2+bx+c=0$, condition for the two roots to be:
$\qquad\quad\quad\;$ (i) real is $D = b^2-4ac\geq 0$
$\qquad\quad\quad\;$ (ii) equal is $D=0$
$\qquad\quad\quad\;$ (iii) positive is $ab<0\;\text{ and } \; ac>0$
$\qquad\quad\quad\;$ (iv) negative is $ab>0\;\text{ and }\; ac>0$
$\qquad\quad\quad\;$ (v) one root positive and one negative is $ac<0$

Prerequisites:
Quadratic Formula (proof)
Sum and product of roots (Proof)

Proof:
(i) Roots of the quadratic equation given by quadratic formula are:

$\qquad\quad \alpha_{1,2} = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

For the roots to be real, the term inside square root should be non negative.

Hence, $D = b^2-4ac\geq 0$

(ii) Difference in both the roots is due to the square root term. Hence for equality $D$ should be equal to $0$.

$\therefore\quad\;\; \alpha_{1,2} = -\dfrac{b}{2a}$

(iii) For both the roots to be positive, both sum and the product of the roots should be positive.

$\therefore\quad\;\; -b/a > 0 \quad\text{ and }\quad c/a > 0$.

Multiplying with $a^2$,

$\Rightarrow\quad\;\; ab < 0 \quad \text{ and }\quad ac > 0$.

(iv) Similarly, for both the roots to be negative, sum of the roots should be negative but the product should be positive.
Hence, $ab>0\;\text{ and }\; ac>0$

(v) Again for one root to be positive and one negative, product of the roots should be negative.
Hence, $ac<0$

Recommended:
Polynomial remainder theorem
Factor Theorem
Sum and product of roots