Copyright © University of Cambridge. All rights reserved.

'Implicitly' printed from

Show menu

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


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.

D&\geq 0&\cr
\Rightarrow (-r)^2 - 4(r^2)(1-r) &\geq& 0 \cr
\Rightarrow r^2 - 4r^2 + 4r^3 &\geq& 0\cr
\Rightarrow 4r^3 &\geq& 3r^2\cr
\Rightarrow r &\geq& 3/4

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.

Steve solved this problem in the following way:

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:

Asymptote at $r = 0$

As $r \to \infty$, $X \to 0$

Part 3:

$$\frac{\mathrm{d}X}{\mathrm{d}r} = \frac{\mathrm{d}}{\mathrm{d}r} \left[ \frac{1}{2r} \left(1 \pm \sqrt{4r - 3} \right) \right] = \frac{-2 \left( \sqrt{4r-3} \right) +4r-6} {r^2 \sqrt{4r-3}}$$

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