Problem source

Find all real values of $x$ that satisfy:
\begin{questionparts}
\item $ \ds \sqrt{3x^2+1} + \sqrt{x} -2x-1=0 \;;$
\item  $ \ds \sqrt{3x^2+1} - 2\sqrt{x} +x-1=0 \;;$
\item  $ \ds \sqrt{3x^2+1} - 2\sqrt{x} -x+1=0 \;.$
\end{questionparts}

Solution source

\begin{questionparts}

\item $\,$ \begin{align*}
&& 0 &= \sqrt{3x^2+1} + \sqrt{x} -2x-1 \\
\Rightarrow && 2x+1 &=  \sqrt{3x^2+1} + \sqrt{x} \\
\Rightarrow && 4x^2+4x+1 &= 3x^2+1+x+2\sqrt{x(3x^2+1)} \\
\Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\
\Rightarrow && x^2(x^2+6x+9) &= 4x(3x^2+1) \\
\Rightarrow && 0 &= x^4-6x^3+9x^2-4x \\
&&&= x(x-4)(x-1)^2
\end{align*}
So clearly we have $x = 0, x = 1, x = 4$. $x = 0$ works, $x = 1$ works, $x = 4$ works.

\item $\,$ \begin{align*}
&& 0 &= \sqrt{3x^2+1} - 2\sqrt{x} +x-1 \\
\Rightarrow && 1-x &= \sqrt{3x^2+1}-2\sqrt{x} \\
\Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\
\Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\
\Rightarrow && 0 &= x(x-4)(x-1)^2
\end{align*}

Again we must check $x = 0, x = 1, x = 4$. $x = 0,1$ work, but $x = 4$ is not a solution.

\item $\,$ \begin{align*}
&& 0 &= \sqrt{3x^2+1} - 2\sqrt{x} -x+1 \\
\Rightarrow && x-1 &= \sqrt{3x^2+1} - 2\sqrt{x} \\
\Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\
\Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)}\\
\Rightarrow && 0 &= x(x-4)(x-1)^2
\end{align*}

So again, we need to check $x = 0, 1, 4$. $x = 0, 4$ work, but $x = 1$ fails.


\end{questionparts}