In this question \(b\), \(c\), \(p\) and \(q\) are real numbers.
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.
Determine necessary and
sufficient conditions for the equation \(\displaystyle x^2 + bx + c = 0\)
to have distinct positive real roots.
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
\(4p^3+ 27q^2\,\).
Solution:
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 \(6 > 0\)
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 \(b^2-4ac > 0, b < 0, 4ac > 0\)
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 \(4p^3+27q^2 < 0\)