Thank you Andrei from Tudor Vianu National College, Romania for your solution.

(1) First, I make the substitution $t = x - 1 / 2$ in the equation:

$2 t^{3} + 3 t^{2} - 11 t - 6 = 0 .$

I obtain:

$2 (x - \frac{1}{2})^{3} + 3 (x - \frac{1}{2})^{2} - 11 (x - \frac{1}{2}) - 6 = 0 .$

This equation is transformed into $x^{3} - \frac{25 x}{4} = 0$ . This equation could be written

$x (x - \frac{5}{2}) (x + \frac{5}{2}) = 0$

with 3 solutions $x = 0, + \frac{5}{2}, - \frac{5}{2}$ which correspond to $t = \frac{1}{2}, 3, - 2$ respectively.

(2) Now I shall analyse what happens to the graph of $y = x^{3} + px + q$ and to the solutions to the equation $x^{3} + px + q = 0$ if I keep $p$ constant and I change $q$ . The graph is translated parallel to the y-axis. There will always be at least one intersection with the x-axis, giving one real root of the equation, and for some values of $q$ this will be the only intersection, for the other values of $q$ there will be 3 intersections, corresponding to 3 real roots and, for one value of $q$ , two of the three with roots are coincident.

(3) If $q$ is fixed, the intersection of the graph with the y-axis is fixed and this time, changing $p$ changes the shape of the graph. Again there is always one real root and, for some values of $p$ there are 3 real roots while for other values of $p$ there is only one real root.

(4) For $p = 0, q = 8$ the applet gives the following solutions to $x^{3} + 8 = 0$ values:

$(- 2, 0) (1, 1.73), (1, - 1.73) .$

Now, I shall solve the equation $x^{3} + 8 = 0$ . I factor the expression as follows:

$(x + 2) (x^{2} - 2 x + 4) = 0 .$

One solution is evidently $x_{1} = - 2$ . The other 2 could be found solving the second order equation:

$x_{2} = 1 + i \sqrt{3}$

$x_{3} = 1 - i \sqrt{3} .$

Now I shall prove that the 3 roots are equally spaced in the Argand diagram, being situated on the circle of centre (0,0) and radius 2. For this, I calculate the moduli of the 3 complex numbers:

$| x_{1} | = 2$

$| x_{2} | = \sqrt{(} 1 + 3) = 2$

$| x_{3} | = \sqrt{(} 1 + 3) = 2 .$

Hence the numbers in modulus argument form $(r, θ)$ are $(2, 0), (2, 120^{o}), (2, 240^{o})$ .

(5) Some results are summarized in the following table:

$p = - 3$ (constant) Red frame (p, q) Blue frame $y = x^{3} + px + q$ Green frame '?? Argand diagram (p, q)

$q < - 2$ (p, q) is on the right of the curve. The graph intersects the x-axis in only one point, with the x-coordinate positive. One point is son the x-axis (x> 0) (corresponding to the real root); the other 2 are symmetric with respect to the x-axis (they are complex conjugates)

$q = - 2$ (p, q) is on the curve. The graph intersects the x-axis in one point and it is an inferior tangent to the x-axis Only real roots, two of which are equal: $x_{1} = 2, x_{2} = x 3 = - 1$

$- 2 < q < 2$ (p, q) is on the left of the curve The graph intersects the x-axis in 3 points, not all having the same sign. All points are on the x-axis.

q = 2 (p, q) is on the curve The graph intersects the x-axis in one point and it is a tangent to the x-axis Only real roots, two of which are equal: $x_{1} = x_{2} = 1, x_{3} = - 2$

$q > 2$ (p, q) is on the right of the curve The graph intersects the x-axis in only one point, with the x-coordinate negative One point is situated on the x-axis (x< 0) (corresponding to the real root); the other 2 are symmetric with respect to the x-axis (they are complex conjugates)

By experimenting with the applet one finds that the curve in the red frame seems to be the boundary such that points $(p, q)$ to the left of this curve correspond to graphs $y = x^{3} + px + q$ with equations $x^{3} + px + q = 0$ having 3 real roots, points on the curve to equations with 3 real roots where 2 are coincident, and points to the right of the curve to equations with only one real root.

The curve has equation $q^{2} = - \frac{4 p^{3}}{27}$ so the points (-3,-2) and (-3,2) are on the curve.

Discriminant The equation

Q (x) = x^{3} + px + q = 0

has derivative

Q' (x) = 3 x^{2} + p

and turning points where

x = \pm \sqrt{-} \frac{p}{3}

. Write

p = - s^{2}

then the turning points are

x = \pm \frac{s}{\sqrt{3}}

. The condition for three real roots is that one turning point has a positive

y

coordinate and the other a negative

y

coordinate, i.e.

[\frac{s^{3}}{3 \sqrt{3}} - \frac{s^{3}}{\sqrt{3}} + q] [- \frac{s^{3}}{3 \sqrt{3}} + \frac{s^{3}}{\sqrt{3}} + q] \leq 0

which reduces to

[q - \frac{2 s^{3}}{3 \sqrt{3}}] [q + \frac{2 s^{3}}{3 \sqrt{3}}] \leq 0

i.e.

q^{2} \leq \frac{4 s^{6}}{27}, q^{2} \leq \frac{4 p^{3}}{27} .

(2)in more detail.Considering the general cubic, the equation

${at}^{3} + {bt}^{2} + ct + d = 0$

becomes

$t^{3} + \frac{b}{a} t^{2} + \frac{c}{a} t + \frac{d}{a} = 0 .$

Making the substitution $t = x - \frac{b}{3 a}$ gives the equation:

$(x - \frac{b}{3 a})^{3} + \frac{b}{a} (x - \frac{b}{3 a})^{2} + \frac{c}{a} (x - \frac{b}{3 a}) + \frac{d}{a} = 0 .$

Expanding and simplifying this equation reduces to:

$x^{3} - 3 (\frac{b}{3 a}) x^{2} + 3 (\frac{b}{3 a})^{2} x - (\frac{b}{3 a})^{3} + \frac{b}{a} (x^{2} - 2 (\frac{b}{3 a}) x + (\frac{b}{3 a})^{2}) + \frac{c}{a} (x - \frac{b}{3 a}) + \frac{d}{a} = 0 .$

Clearly the coefficient of $x^{2}$ is zero and the equation reduces to $x^{3} + px + q = 0$ where $p$ and $q$ depend on $a, b, c$ and $d$ .

$p = - 3$ (constant)	Red frame (p, q)	Blue frame $y = x^{3} + px + q$	Green frame '?? Argand diagram (p, q)
$q < - 2$	(p, q) is on the right of the curve.	The graph intersects the x-axis in only one point, with the x-coordinate positive.	One point is son the x-axis (x> 0) (corresponding to the real root); the other 2 are symmetric with respect to the x-axis (they are complex conjugates)
$q = - 2$	(p, q) is on the curve.	The graph intersects the x-axis in one point and it is an inferior tangent to the x-axis	Only real roots, two of which are equal: $x_{1} = 2, x_{2} = x 3 = - 1$
$- 2 < q < 2$	(p, q) is on the left of the curve	The graph intersects the x-axis in 3 points, not all having the same sign.	All points are on the x-axis.
q = 2	(p, q) is on the curve	The graph intersects the x-axis in one point and it is a tangent to the x-axis	Only real roots, two of which are equal: $x_{1} = x_{2} = 1, x_{3} = - 2$
$q > 2$	(p, q) is on the right of the curve	The graph intersects the x-axis in only one point, with the x-coordinate negative	One point is situated on the x-axis (x< 0) (corresponding to the real root); the other 2 are symmetric with respect to the x-axis (they are complex conjugates)