Implicitly

Two of our best solvers, Alex from Stoke on Trent Sixth Form College and Patrick from Woodbridge School thought about this implicit equation.

$$r^2{X}^2-rX-r+1=0$$

Patrick looked at the first parts using the quadratic equation formula as follows:

Part 1:

If $r=1$ then $X^2-X = 0$, so $X(X-1) = 0$ and $X = 0$ or $1$.

If $r = 100$ then $100^2X^2 - 100X - 100 + 1 = 0$ so $100^2X^2-100X-99 = 0$ so

$$X=\frac{100\pm \sqrt{100²+4\times 100²\times 99}}{2\times {100}^2}$$

This is real as we neither divide by $0$ nor take a negative square root.

Part 2:

If $r = 0$ then we find the equation gives $1=0$, so r cannot equal $0$.

Substituting $r^2$ for $a$, $-r$ for $b$ and $-(r-1)$ for $c$ in the usual quadratic equation formula we get

$$X=\frac{r\pm \sqrt{r²+4r²(r-1)}}{2r^2}=\frac{1\pm \sqrt{1+4r-4}}{2r}=\frac{1\pm \sqrt{4r-3}}{2r}$$

So this takes real values if and only if $4r - 3 \geq 0$.

Alex realised that the key feature of this problem was the discriminant of the quadratic equations in $X$, and looked at that object directly:

The discriminant determines whether a quadratic has real roots, therefore $X(r)$ will have real values if and only if the discriminant $D$ of the quadratic is zero or positive.

$$\begin{array}{rcl}D& \ge 0& \\ \Rightarrow (-r{)}^2-4(r^2)(1-r)& \ge & 0\\ \Rightarrow r^2-4r^2+4r^3& \ge & 0\\ \Rightarrow 4r^3& \ge & 3r^2\\ \Rightarrow r& \ge & 3/4\end{array}$$

So, $X(r)$ has real values for $r\geq 3/4$.

This means that $X(1)$ and $X(100)$ have real values, but $X(-1)$ does not. $X(0)$ is undefined because the corresponding equation $(1=0)$ has no solutions, it is a contradiction.

Alex also correctly deduced the maximum and minimum values of $X(r)$

The maximum value of $X(r)$ is $X(1) = 1$. The minimum value of $X(r)$ is $X(3) = -1/3$. The graph has an asymptote $r = 0$.

Alex submitted a plot of the associated curve, which shows two distinct branches.

Image

Part 1:

$$X=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$X=\frac{1\pm \sqrt{4r-3}}{2r}$$

$X$ is therefore real for all $r$ greater than or equal to $\frac{3}{4}$

Hence $r = 0$ and $-1$ give complex roots of $X$ but $1$ and $100$ give real roots.

Part 2:

Image

Part 3:

$$\frac{\mathrm{d}X}{\mathrm{d}r}=\frac{\mathrm{d}}{\mathrm{d}r}\left[\frac{1}{2r}(1\pm \sqrt{4r-3})\right]=\frac{-\sqrt{4r-3}\pm (3-2r)}{2r^2\sqrt{4r-3}}$$

Setting $\displaystyle \frac{\mathrm{d}X}{\mathrm{d}r} = 0$ we find $r = 1,\ 3$

Editor's Note: The Maximum value of $X(r)$ occurs when $r=1$ and we take the upper branch of the curve (corresponding to the positive square root taken in the quadratic formula).  The minimum value of $X(r)$ occurs when $r=3$ in the lower branch.

This gives $X_{max}=1$ and $X_{min}=-\frac 13$.