Find the points on the $x$-axis which are at a distance of $2\sqrt{5}$ from the point $(7, -4)$. How many such points are there?

(A) We know that every point on the $x$-axis is of the form $(x, 0)$. Let $P(x, 0)$ be a point on the $x$-axis at a distance of $2\sqrt{5}$ from the point $Q(7, -4)$.
Using the distance formula,$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Given $PQ = 2\sqrt{5}$,so $(PQ)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20$.
Substituting the coordinates:
$(x - 7)^2 + (0 - (-4))^2 = 20$
$(x - 7)^2 + (4)^2 = 20$
$x^2 - 14x + 49 + 16 = 20$
$x^2 - 14x + 65 = 20$
$x^2 - 14x + 45 = 0$
Factoring the quadratic equation:
$x^2 - 9x - 5x + 45 = 0$
$x(x - 9) - 5(x - 9) = 0$
$(x - 9)(x - 5) = 0$
Thus,$x = 5$ or $x = 9$.
The points are $(5, 0)$ and $(9, 0)$.
There are $2$ such points.