, 2014

We note that $BD = BC+CD$ and that $BC = 20$ cm, so we need to determine $CD$.

We draw a line from $C$ to $P$ on $FD$ so that $CP$ is perpendicular to $DF$.

Since $AC$ and $DF$ are parallel, then $CP$ is also perpendicular to $AC$.

The distance between $AC$ and $DF$ is 4 cm, so $CP = 4$ cm.

Since $\triangle ABC$ is isosceles and right-angled, then $\angle ACB = 45^\circ$.

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Thus, $\angle PCD = 180^\circ - \angle ACB - \angle PCA = 180^\circ - 45^\circ - 90^\circ = 45^\circ$.

Since $\triangle CPD$ is right-angled at $P$ and $\angle PCD = 45^\circ$, then $\triangle CPD$ is also an isosceles right-angled triangle.

Therefore, $CD = \sqrt{2}CP = 4\sqrt{2}$ cm.

Finally, $BD = BC + CD = (20+4\sqrt{2})$ cm.
Manipulating the given equation and noting that $x\neq 0$ and $x \neq -\frac{1}{2}$ since neither denominator can equal 0, we obtain $$\begin{aligned} \dfrac{x^2+x+4}{2x+1} & = \dfrac{4}{x}\\ x(x^2+x+4) & = 4(2x+1) \\ x^3 + x^2 + 4x & = 8x + 4 \\ x^3 + x^2 - 4x - 4 & = 0 \\ x^2(x+1)-4(x+1) & = 0 \\ (x+1)(x^2-4) & = 0 \\ (x+1)(x-2)(x+2) & = 0\end{aligned}$$ Therefore, $x=-1$ or $x=2$ or $x=-2$. We can check by substitution that each satisfies the original equation.