My 11th grade high school physics teacher gave me a 1st year bachelor level question which can be solved using 11th grade physics and math.
Imagine an aluminum pendulum.
It is swinging from point $O$ with length $l$ while the angle between the pendulum and the norm is $\theta$. The distance between the norm and the two metal sheets is the same, such that the distance between the two metal sheets is $d$.
Let the electrical charge of the pendulum be $q_{1}$, and the charges of the two metal sheets be $q_{2}$ and $q_{3}$.
$q_{2}$ is greater than $q_{3}$, and all charges $q_{1}$, $q_{2}$ and $q_{3}$ are positive, so there is an angle where the pendulum is constant, meaning the two electromagnetic forces on the pendulum are the same.
Question: Let $x = sin(\theta)$. Find a second degree polynomial and show that one of the two solutions is mathematically impossible. Then express $\theta$ in terms of $d$, $l$, $q_{2}$ and $q_{3}$.
Here's my answer:
Let $d_{\theta}$ the horizontal distance between the vertical $O$ and the pendulum, such that \begin{equation} \sin{\theta} = \frac{d_{\theta}}{l} \end{equation}
and
\begin{equation} \begin{split} \sin{\theta} &= x = \frac{d_{\theta}}{l} \\ d_{\theta} &= lx \end{split} \end{equation}
Also, let $d_{1/2}$ the horizontal distance between the leftmost metal sheet and the pendulum and $d_{1/3}$ that between the pendulum and the rightmost metal sheet so that we have the following: \begin{eqnarray} d_{1/2} = \frac{d}{2} + d_{\theta} \\ d_{1/3} = \frac{d}{2} - d_{\theta} \end{eqnarray} such that $d_{1/2} > d_{1/3}$ meaning $|q_{2}| > |q_{3}|$.
Since the pendulum is static, the electromagnetic forces $F_{q_{1}/q_{2}}$ and $F_{q_{1}/q_{3}}$ exerted from the two metal sheets are equal \begin{equation} \label{eq:sol} F_{q_{1}/q_{2}} = F_{q_{1}/q_{3}} \end{equation} and according to Coulomb's law: \begin{eqnarray*} F_{q_{1}/q_{2}} = k\times \frac{|q_{1}|\times |q_{2}|}{{d_{1/2}}^{2}} \\ F_{q_{1}/q_{3}} = k\times \frac{|q_{1}|\times |q_{3}|}{{d_{1/3}}^{2}} \end{eqnarray*}
We must therefore solve our electromagnetic forces equality. \begin{align*} F_{q_{1}/q_{2}} &= F_{q_{1}/q_{3}} \\ k\times \frac{|q_{1}|\times |q_{2}|}{{d_{1/2}}^{2}} &= k\times \frac{|q_{1}|\times |q_{3}|}{{d_{1/3}}^{2}} \\ \frac{|q_{1}|\times |q_{2}|}{{d_{1/2}}^{2}} &= \frac{|q_{1}|\times |q_{3}|}{{d_{1/3}}^{2}} \\ \frac{|q_{2}|}{{d_{1/2}}^{2}} &= \frac{|q_{3}|}{{d_{1/3}}^{2}} \\ \frac{{d_{1/2}}^{2}}{{d_{1/3}}^{2}} &= \frac{|q_{2}|}{|q_{3}|} \end{align*}
We plug in the definitions of $d_{1/2}$ and $d_{1/3}$ to get
\begin{equation} \begin{split} \frac{(\frac{d}{2} + d_{\theta})^{2}}{(\frac{d}{2} - d_{\theta})^{2}} &= \frac{|q_{2}|}{|q_{3}|} \\ |q_{3}|(\frac{d}{2} + d_{\theta})^{2} &= |q_{2}|(\frac{d}{2} - d_{\theta})^{2} \\ |q_{3}|(\frac{d}{2} + d_{\theta})^{2} - |q_{2}|(\frac{d}{2} - d_{\theta})^{2} &= 0 \\ |q_{3}|({d_{\theta}}^{2} + d_{\theta}d + \frac{d^2}{4}) - |q_{2}|({d_{\theta}}^{2} - d_{\theta}d + \frac{d^2}{4}) &= 0 \\ |q_{3}|{d_{\theta}}^{2} + |q_{3}|d_{\theta}d + |q_{3}|\frac{d^2}{4} - (|q_{2}|{d_{\theta}}^{2} - |q_{2}|d_{\theta}d + |q_{2}|\frac{d^2}{4}) &= 0 \\ |q_{3}|{d_{\theta}}^{2} + |q_{3}|d_{\theta}d + |q_{3}|\frac{d^2}{4} - |q_{2}|{d_{\theta}}^{2} + |q_{2}|d_{\theta}d - |q_{2}|\frac{d^2}{4} &= 0 \\ (|q_{3}| - |q_{2}|){d_{\theta}}^{2} + [d(|q_{3}| + |q_{2}|)]d_{\theta} + \frac{d^2(|q_{3}| - |q_{2}|)}{4} &= 0 \end{split} \end{equation}
We insert $d_{\theta} = lx$ to obtain the following second degree polynomial equation \begin{equation} \begin{split} (|q_{3}| - |q_{2}|){l}^{2}{x}^{2} + d(|q_{3}| + |q_{2}|)lx + \frac{d^2(|q_{3}| - |q_{2}|)}{4} &= 0 \\ {l}^{2}(|q_{3}| - |q_{2}|){x}^{2} + dl(|q_{3}| + |q_{2}|)x + \frac{d^2(|q_{3}| - |q_{2}|)}{4} &= 0 \end{split} \end{equation} with $a = {l}^{2}(|q_{3}| - |q_{2}|)$, $b = dl(|q_{3}| + |q_{2}|)$, and $c = \frac{d^2(|q_{3}| - |q_{2}|)}{4}$.
We determine its discriminant in order to solve it. \begin{equation} \begin{split} \Delta &= b^{2} - 4ac \\ &= [dl(|q_{3}| + |q_{2}|)]^{2} - 4\times {l}^{2}(|q_{3}| - |q_{2}|)\times \frac{d^2(|q_{3}| - |q_{2}|)}{4} \\ &= d^{2}l^{2}[(|q_{3}| + |q_{2}|)^{2} - d^2{l}^{2}(|q_{3}| - |q_{2}|)^{2} \\ &= d^{2}l^{2}[(|q_{3}| + |q_{2}|)^{2} - (|q_{3}| - |q_{2}|)^{2}] \\ &= d^{2}l^{2}[(|q_{3}| + |q_{2}| + |q_{3}| - |q_{2}|)(|q_{3}| + |q_{2}| - |q_{3}| + |q_{2}|)] \\ &= d^{2}l^{2}|q_{3}|^{2}|q_{2}|^{2} \\ \end{split} \end{equation}
$\Delta > 0$, so there exists two solutions.
\begin{equation*} \begin{aligned}[c] x_{1} &= \frac{\sqrt{\Delta} - b}{2a}\\ x_{1} &= \frac{\sqrt{d^{2}l^{2}|q_{3}|^{2}|q_{2}|^{2}} - dl(|q_{3}| + |q_{2}|)}{2{l}^{2}(|q_{3}| - |q_{2}|)}\\ x_{1} &= \frac{dl|q_{3}||q_{2}| - dl(|q_{3}| + |q_{2}|)}{2{l}^{2}(|q_{3}| - |q_{2}|)}\\ x_{1} &= \frac{d(|q_{3}||q_{2}| - |q_{3}| - |q_{2}|)}{2l(|q_{3}| - |q_{2}|)} \end{aligned} \qquad\qquad \begin{aligned}[c] x_{2} &= \frac{-\sqrt{\Delta} - b}{2a}\\ x_{2} &= \frac{-\sqrt{d^{2}l^{2}|q_{3}|^{2}|q_{2}|^{2}} - dl(|q_{3}| + |q_{2}|)}{2{l}^{2}(|q_{3}| - |q_{2}|)}\\ x_{2} &= \frac{-dl|q_{3}||q_{2}| - dl(|q_{3}| + |q_{2}|)}{2{l}^{2}(|q_{3}| - |q_{2}|)}\\ x_{2} &= \frac{-d(|q_{3}||q_{2}| - |q_{3}| - |q_{2}|)}{2l(|q_{3}| - |q_{2}|)} \end{aligned} \end{equation*}
Now I've reached an obstacle. I have to prove that one of the two solutions $x_{1}$ and $x_{2}$ is in $[-1 ; 1]$ since it's an image of $\sin$, and the image of that solution by $\arcsin$ would be $\theta$.
Your discriminant calculation is wrong. Whenever you have something of the form $(a+b)^2-(a-b)^2$ it should simplify down like this:
$$ (a+b)^2-(a-b)^2 = a^2 +2ab+b^2-(a^2-2ab+b^2) = 4ab$$
So your discriminant would really be $\Delta = 4d^2l^2q_3q_2$
Then take a look at your solutions. They both end up being perfect squares in the numerator.
$$x_1 = \frac{-d(q_3-2\sqrt{q_3q_2} + q_2)}{2l(q_3-q_2)} = -\frac{d}{2l}\frac{(\sqrt{q_3}-\sqrt{q_2})^2}{(\sqrt{q_3}-\sqrt{q_2})(\sqrt{q_3}+\sqrt{q_2})} = \frac{d}{2l}\frac{\sqrt{q_2}-\sqrt{q_3}}{\sqrt{q_2}+\sqrt{q_3}}$$
$$x_2 = \frac{-d(q_3+2\sqrt{q_3q_2} + q_2)}{2l(q_3-q_2)} = -\frac{d}{2l}\frac{(\sqrt{q_3}+\sqrt{q_2})^2}{(\sqrt{q_3}-\sqrt{q_2})(\sqrt{q_3}+\sqrt{q_2})} = \frac{d}{2l}\frac{\sqrt{q_2}+\sqrt{q_3}}{\sqrt{q_2}-\sqrt{q_3}}$$
Then you have
$$x_1 < \frac{d}{2l} < x_2$$
so pick whichever solution works best for you in the context of the problem.