Problem source

In this question $b$, $c$, $p$ and $q$ are real numbers.
\begin{questionparts}
\item By considering the graph $y=x^2 + bx + c$ 
show that $c < 0$ is a sufficient condition for the equation 
$\displaystyle x^2 + bx + c = 0$ to have distinct real roots. 
Determine whether $c < 0$ is a necessary condition for the 
equation to have distinct real roots.

\item
Determine necessary and 
sufficient conditions for the equation $\displaystyle x^2 + bx + c = 0$
to have  distinct positive real roots.

\item What can be deduced about the 
number and the nature of the roots of the equation 
$x^3 + px + q = 0$ if $p>0$ and $q<0$?
What can be deduced if $p<0\,$ and $q<0$? You should consider 
the different cases that arise according to the value of
$4p^3+ 27q^2\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)*(#1)+2*(#1)-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)});
    \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}

Since $y(0) < 0$ and $y(\pm \infty) > 0$ we must cross the axis twice. Therefore there are two distinct real roots.

It is not necessary, for example $(x-2)(x-3)$ has distinct real roots by the constant term is $6 > 0$

\item For $x^2+bx+c=0$ to have distinct, positive real roots we need $\Delta > 0$ and $\frac{-b -\sqrt{\Delta}}{2a} > 0$ where $\Delta = b^2-4ac$, ie $b < 0$ and $b^2 > \Delta = b^2-4ac$ or $4ac > 0$. Therefore we need $b^2-4ac > 0, b < 0, 4ac > 0$

\item Since $q < 0$ at least one of the roots is positive. The gradient is $3x^2+p > 0$ therefore there is exactly one positive root.

If $p < 0$ then there are turning points when $3x^2+p = 0$ ie $x = \pm \sqrt{\frac{-p}{3}}$. If the first turning point is above the $x$-axis then there will be 3 roots. If it is on the $x$-axis then 2, otherwise only 1.

\begin{align*}
y &= \left (-\sqrt{\frac{-p}{3}}\right)^3 + p\left (-\sqrt{\frac{-p}{3}}\right)+q \\
&= \sqrt{\frac{-p}{3}} \left (p - \frac{p}{3} \right) + q \\
&= \frac{2}{3} \sqrt{\frac{-p}{3}}p +q \\
\end{align*}

Therefore it is positive if $-\frac{4}{27}p^3 >q^2$ ie if $4p^3+27q^2 < 0$
\end{questionparts}