Why sometimes we get only one root of quadratic equations? What is logic behind getting (sometimes) only one root of a quadratic equation which satisfies the equation?
 A: The logic is that every positive number has two square roots, but zero only has one.
The quadratic equation comes from the algebraic process known as completing the square.  Given an equation $ax^2 + bx + c = 0$, divide by $a$:
$$
x^2 + \frac{b}{a}x + \frac{c}{a} = 0
$$
Then add $\left(\frac{b}{2a}\right)^2$ and to both sides:
$$
x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = \left(\frac{b}{2a}\right)^2
$$
Then subtract $\frac{c}{a}$:
$$
x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a} 
$$
The left-hand side is a perfect square (by design).  We rearrange the right-hand side at the same time:
$$
\left(x + \frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}
$$
At this point we would take the square root of both sides.  Since the denominator is positive, the number of roots depends only on the numerator.  If the numerator is positive, there are two roots.  If the number is negative, there are zero (real) roots.  If the numerator is zero there is only one.
Now you see why $b^2-4ac$ is called the discriminant of a quadratic equation.  It detects the number of roots (or the type of roots if you allow complex ones).  To finish up, we have
$$
x+\frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}
$$
so
$$
x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
$$
A: The vertex of your parabola is ON the $x$-axis.  So it only touches it once.

A: The situation for a quadratic function and its roots is roughly like this

A: If a root is repeated, there are two non-disctinct roots. Discriminant $ b^2- 4 a c$ vanishes.
A: Let $ax^2 + bx + c = 0$, with $a \neq 0$, be a quadratic equation with real coefficients.  Then the roots are given by the Quadratic Formula
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The quantity inside the square root is called the discriminant, which is denoted $\Delta$.  If 


*

*$\Delta > 0$, there are two distinct real roots, namely
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
In this case, the graph has two distinct $x$-intercepts.

*$\Delta = 0$, there is one double root, namely
$$x = -\frac{b}{2a}$$
In this case, the polynomial factors as 
$$a\left(x + \frac{b}{2a}\right)^2 = 0$$
so the two roots coincide.  In this case, the parabola is tangent to the $x$-axis.

*$\Delta < 0$, there are two complex conjugate roots, namely
$$x = \frac{-b \pm i\sqrt{|\Delta|}}{2a}$$
In this case, the graph does not intersect the $x$-axis.
