Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

Target Six

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

8 Methods for Three by One

This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

Root Tracker

Stage: 5 Challenge Level:

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 $$z_1z_2 = {-p + \sqrt {p^2 - 4q}\over 2}\times {-p^2 - \sqrt {p^2 - 4q}\over 2} = {p^2 - (p^2 - 4q)\over 4} = q.$$