Problem source

Let 
\[\f(x) = a \sqrt{x} - \sqrt{x - b}\;,
\]
 where $x\ge b >0$ and $a>1\,$.
Sketch the graph of $\f(x)\,$.
Hence show that the equation $\f(x)  = c$, where $c>0$, 
has no solution when $c^2 < b \l a^2 - 1 \r\,$.
Find conditions on $c^2$ in terms of $a$ and $b$ 
for the equation to have exactly one or exactly two solutions.
Solve the equations 
\begin{questionparts}
	\item  $3 \sqrt{x} - \sqrt{x - 2} = 4\, ,$ 
\item   $3 \sqrt{x} - \sqrt{x - 3} = 5\;$.
\end{questionparts}

Solution source

\begin{align*}
&& f'(x) &= \frac12  ax^{-1/2}-\frac12(x-b)^{-1/2} \\
\Rightarrow f'(x) = 0: && 0 &= \frac{a\sqrt{x-b}-\sqrt{x}}{\sqrt{x(x-b)}} \\
\Rightarrow && x &= a^2(x-b)\\ 
\Rightarrow && x &= \frac{a^2b}{a^2-1}  \\
&& f(x) &= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{a^2b}{a^2-1}-b} \\
&&&= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{b}{a^2-1}} \\
&&&= \sqrt{b(a^2-1)}
\end{align*}


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1.5*sqrt(#1)-sqrt((#1)-1)};
    \def\xl{-1};
    \def\xu{15};
    \def\yl{-1};
    \def\yu{5};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=1:\xu, samples=201] 
            plot (\x, {\functionf(\x)});
        \draw[thin, red, dashed, domain=0:\xu, samples=201] 
            plot (\x, {1.5*sqrt(\x)});
        \draw[thin, red, dashed, domain=1:\xu, samples=201] 
            plot (\x, {1*sqrt(\x-1)});
        \draw[thin, red, dashed, domain=0:\xu, samples=201] 
            plot (\x, {0.5*sqrt(\x)});
    \end{scope}

    \filldraw ({1.5*1.5/(1.5*1.5-1)},{sqrt(1.5*1.5-1)}) circle (1.5pt) node[below] {$^{\left (\frac{a^2b}{a^2-1}, \sqrt{b(a^2-1)}\right)}$};
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}


If $c$ is below the turning point, ie $c^2 < b(a^2-1)$ there is no solution.

If $c^2 = b(a^2-1)$ there is exactly one solution.

If $b(a^2-1) < c^2 < (f(b))^2 = a^2b$ then there are two solutions, otherwise there is exactly one solution.

\begin{questionparts}
\item $c^2 = 16$, $2 \cdot (3^2-1) = 16$, so we should have exactly one solution at $x = \frac{3^2 \cdot 2}{3^2 -1 } = \frac{9}{4}$

\item $c^2 = 25$ and $3 \cdot (3^2 - 1) = 24, 3 \cdot (3^2) = 27$, so we look for two solutions.

\begin{align*}
&& 5 & = 3 \sqrt{x} - \sqrt{x-3}  \\
\Rightarrow && 25 &= 9x+x-3-6\sqrt{x(x-3)} \\
\Rightarrow && 3\sqrt{x(x-3)} &= 5x-14 \\
\Rightarrow && 9x(x-3) &= 25x^2-140x+196 \\
\Rightarrow && 0 &= 16x^2-113x+196 \\
&&&= (x-4)(16x-49) \\
\Rightarrow && x &= 4, \frac{49}{16}
\end{align*}

\end{questionparts}