Tiffany from Island School and Andrei from Tudor Vianu National College, Romania sent in very good solutions.

(1) If you keep $p$ constant and change $q$, the graph of $$y=x^2 + px + q \quad (1)$$ is translated parallel to the y axis. Consequently the number of intersections of the graph with the x axis (i.e. of real solutions of the equation $x^2+px+q=0$) changes. This number is related to the discriminant of the equation: $\Delta = p^2 -4q \quad (2)$.

- if $\Delta > 0$ there are 2 real, distinct solutions

- if $\Delta = 0$ there is a repeated (or double) real solution

- if $\Delta < 0$ there are no real solutions.

a) For $p=-5, q=-6$: $\Delta = 49$ and the x-intercepts are (-1,0) and (6,0) showing 2 real solutions to the equation .

b) For $p= -5, q= 4$: $\Delta = 9$ and the x-intercepts are (1,0) and (4,0) showing 2 real solutions.

c) For $p = -5, q = 7$: $\Delta = -3$ and there are no intersection with the x axis and no real solutions.

(2) If you keep $q$ constant and change $p$ the intersection of the graph of $y=x^2 + px + q$ with the y-axis is kept fixed at the point $(0,q)$. The number of intersections with the x-axis varies and it could be 0, 1 or 2 depending again on $\Delta$.

From relation (2) it is easy to observe that $\Delta$ has the same value if p changes sign and that for $p^2 > 4q$ and for $p^2 < -4q$ there are 2 distinct real solutions, for $p^2 = 4q$ there are two equal real solutions and for the rest there are no real solutions.

a) For $p=-10, q = 16$ the solutions are $x=2$ and $x=8$.

b) For $p=-8, q=16$ the graph is tangent to the x-axis and there is a repeated real solution $x=4$.

c) $p= -6,\ q=16$ there are no x-intercepts and we shall see that there are 2 complex conjugate solutions.

(3) and (4)

When the point $(p,q)$ is in the region below the parabola $p^2 = 4q$, showing $\Delta = p^2 - 4q > 0$, the graph of $y = x^2 + px + q$ crosses the x-axis in two distinct points and the equation $x^2 + px + q = 0$ has 2 distinct real solutions. In this case the the roots show up on the $u$-axis in the Argand Diagram (called the real axis ).

When the point $(p,q)$ enters the region above the parabola $p^2 = 4q$ the graph of $y = x^2 + px + q$ no longer crosses the x-axis and the equation $x^2 + px + q = 0$ has no real solutions. In this region the movement of the point $(p,q)$ in the red frame leaves a red track showing $\Delta = p^2 - 4q < 0$.

When the point $(p,q)$ is on the boundary between the two regions, that is on the parabola $p^2 = 4q$, the graph of $y = x^2 + px + q$ is tangent to the x-axis, there are 2 equal real solutions shown by two coincident points on the $u$-axis in the Argand diagram.

The roots show up on the $v$-axis in the Argand Diagram (called the imaginary axis ) when $p=0$ and $q> 0$.

(5) The roots of the equation $x^2 -6x +13 = 0$ are the complex numbers: $z_1 = 3 + 2i$ and $z_2 = 3 - 2i$ where $i = \sqrt {-1}$.

They satisfy the Vi\'{e}te relations $z_ 1 + z_2 =6$ and $z_1 \times z_2 = 13$. Checking that $z_1^2 -6z_1 + 13 = 0 $ and $z_2^2 -6z_2 + 13 = 0 $ we verify that these are solutions of the equation.

(7) Two complex roots of the quadratic equation $x^2 + px +q = 0$ (where $p$ and $q$ are real numbers and $p^2 < 4q$) are $$z_1 = {-p + \sqrt {p^2 - 4q}\over 2} = u_1 + iv_1 \quad {\rm and}\quad z_2 = {-p - \sqrt {p^2 - 4q}\over 2} = u_2 + iv_2.$$

We see that $u_1 = u_2 = {-p\over 2}$ and $v_1 = - v_2 = \sqrt {p^2 - 4q}$ so the points in the Argand diagram representing these solutions are reflections of each other in the real axis.

The sum of the roots is $z_1 + z_2 = p$.

The product of the roots is $$z_1z_2 = {-p + \sqrt {p^2 - 4q}\over 2}\times {-p^2 - \sqrt {p^2 - 4q}\over 2} = {p^2 - (p^2 - 4q)\over 4} = q. $$