Problem source

Solve  the inequalities
\begin{questionparts}
\item $1+2x-x^2 >2/x \quad (x\ne0)$ ,
\item $\sqrt{3x+10} > 2+\sqrt{x+4} \quad (x\ge -10/3)$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){-(#1)^2+2*(#1)+1};
    \def\functiong(#1){2/(#1)};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-10};
    \def\yu{10};
    
    % 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=\xl:\xu, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, red, smooth, domain=\xl:-0.01, samples=100] 
            plot (\x, {\functiong(\x)});
        \draw[thick, red, smooth, domain=0.01:\xu, samples=100] 
            plot (\x, {\functiong(\x)});
    \end{scope}
    
    % 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}
\begin{align*}
&& 1+2x-x^2 = 2/x \\
\Rightarrow && 0 &= x^3-2x^2-x+2 \\
&&&= (x+1)(x^2-3x+2) \\
&&&= (x+1)(x-1)(x-2)
\end{align*}

Therefore the inequality is satisfied on $(1,2)$ and $(-1,0)$

\item $\,$
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sqrt(3*(#1)+10)};
    \def\functiong(#1){2+sqrt((#1)+4)};
    \def\xl{-4};
    \def\xu{20};
    \def\yl{-1};
    \def\yu{20};
    
    % 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={-10/3}:\xu, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, red, smooth, domain=\xl:-0.01, samples=100] 
            plot (\x, {\functiong(\x)});
        \draw[thick, red, smooth, domain=0.01:\xu, samples=100] 
            plot (\x, {\functiong(\x)});
    \end{scope}
    
    % 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}

\begin{align*}
&& \sqrt{3x+10} &= 2+\sqrt{x+4} \\
&& 3x+10 &= x+8 + 4\sqrt{x+4} \\
&& 16(x+4) &= 4(x+1)^2 \\
&& 4x+16 &= x^2+2x+1 \\
\Rightarrow && 0 &= x^2-2x-15 \\
 &&&= (x-5)(x+3)
\end{align*}

Therefore $x > 5$
\end{questionparts}