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2006 Paper 1 Q3
D: 1500.0 B: 1500.0

In this question \(b\), \(c\), \(p\) and \(q\) are real numbers.

  1. 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.
  2. Determine necessary and sufficient conditions for the equation \(\displaystyle x^2 + bx + c = 0\) to have distinct positive real roots.
  3. 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:

  1. TikZ diagram
    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 \(6 > 0\)
  2. 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 \(b^2-4ac > 0, b < 0, 4ac > 0\)
  3. 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 \(4p^3+27q^2 < 0\)