(1) If you keep $p$ constant and change $q$, the graph of

${\displaystyle y=x^2+\mathrm{px}+q\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)}$

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 $x^2+\mathrm{px}+q=0$) changes. This number is related to the discriminant of the equation: $\Delta =p^2-4q\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)$.

- 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 $y=x^2+\mathrm{px}+q$ 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) $p=-6,\hspace{0.5em}q=16$ 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 $y=x^2+\mathrm{px}+q$ crosses the x-axis in two distinct points and the equation $x^2+\mathrm{px}+q=0$ 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 $y=x^2+\mathrm{px}+q$ no longer crosses the x-axis and the equation $x^2+\mathrm{px}+q=0$ 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 $y=x^2+\mathrm{px}+q$ 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 Viéte 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 $x^2+\mathrm{px}+q=0$ (where $p$ and $q$ are real numbers and $p^2<4q$) are

${\displaystyle z_1=\frac{-p+\sqrt{p^2-4q}}{2}=u_1+{\mathrm{iv}}_1\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{and}\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}{z}_2=\frac{-p-\sqrt{p^2-4q}}{2}=u_2+{\mathrm{iv}}_2.}$

We see that $u_1=u_2=\frac{-p}{2}$ 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

${\displaystyle z_1{z}_2=\frac{-p+\sqrt{p^2-4q}}{2}\times \frac{-p^2-\sqrt{p^2-4q}}{2}=\frac{p^2-(p^2-4q)}{4}=q.}$