hard algebra question | Racquel Sanderson

Find the value(s) of $k$ such that the equation below has two numerically equal but opposite sign solutions (e.g. $5$ and $-5$ ).

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{k-1}{k+1}\end{equation}$

$\begin{equation*}(x^2-2x)(k+1)=(k-1)(4x-1)\end{equation}$

$\begin{equation*}(k+1)x^2-2kx-2x=4kx-k-4x+1\end{equation}$

$\begin{equation*}(k+1)x^2-2kx-4kx-2x+4x-1=0\end{equation}$

$\begin{equation*}(k+1)^2x^2-(6k-2)x-1=0\end{equation}$

For there to be two numerically equal but opposite sign solutions, the $b$ term of the quadratic equation must be $0$ .

$\begin{equation*}6k-2=0\end{equation}$

Hence $k=\frac{1}{3}$ .

When $k=\frac{1}{3}$ the equation becomes

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{\frac{-2}{3}}{\frac{4}{3}}\end{equation}$

$\begin{equation*}\frac{x^2-2x}{4x-1}=\frac{-1}{2}\end{equation}$

$\begin{equation*}2x^2-4x=-4x+1\end{equation}$

$\begin{equation*}2x^2-1=0\end{equation}$

$\begin{equation*}x^2=\frac{1}{2}\end{equation}$

$\begin{equation*}x=\pm \frac{1}{\sqrt{2}}\end{equation}$

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