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{\bf
Problem 4.12
}
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Prove that $(1-3x) = \sqrt{5x-1}$ implies $x = 2/9$

\begin{proof}
Suppose $x$ satisfies $(1-3x) = \sqrt{5x-1}$. Squaring both sides yields
\begin{eqnarray*}
 5x-1 &=& (1-3x)^2 \\
5x-1&=& 1-6x+9x^2\\
0 &=&2-11x+9x^2.
\end{eqnarray*}
Using the quadratic formula gives $$x = \frac{11 \pm \sqrt{11^2-4\cdot2\cdot9}}{2\cdot 9} = \frac{11 \pm 7}{18} = 1, \frac{2}{9}.$$
However, the case of $x=1$ must be thrown out since $$(1-3\cdot1) = -2 \neq 2 = \sqrt{5\cdot 1-1}.$$ Hence $x=\frac{2}{9}$.
\end{proof}

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