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2006 Paper 1 Q3
D: 1500.0 B: 1500.0
proof quadratic discriminant sufficient condition necessary condition curve sketching cubic roots of polynomials inequality

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 \(4p^3+27q^2 < 0\)

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2001 Paper 1 Q3
D: 1500.0 B: 1516.0
polynomials curve sketching cubic discriminant inequality roots of polynomials symmetric functions sufficient condition

Sketch, without calculating the stationary points, the graph of the function \(\f(x)\) given by \[ \f(x) = (x-p)(x-q)(x-r)\;, \] where \(p < q < r\). By considering the quadratic equation \(\f'(x)=0\), or otherwise, show that \[ (p+q+r)^2 > 3(qr+rp+pq)\;. \] By considering \((x^2+gx+h)(x-k)\), or otherwise, show that \(g^2>4h\,\) is a sufficient condition but not a necessary condition for the inequality \[ (g-k)^2>3(h-gk) \] to hold.

Solution:

TikZ diagram
Since there are two turning points the derivative (a quadratic) has two distinct real roots. \begin{align*} && f'(x) &= 3x^2-2(p+q+r)x+(pq+qr+rp) \\ && 0 &< \Delta = 4(p+q+r)^2 - 4\cdot 3(pq+qr+rp) \\ \Rightarrow && (p+q+r)^2 &> 3(pq+qr+rp) \end{align*} If \(g^2 > 4h\) then \(p(x) = (x^2+gx+h)(x-k)\) has at least 2 real roots (possibly one repeated, and in particular it has two turning point, ie \begin{align*} && p'(x) &= (2x+g)(x-k)+(x^2+gx+h) \\ &&&= 3x^2+(2g-2k)x + (h-kg) \\ && 0 &< \Delta = 4(g-k)^2 - 4\cdot 3 (h-gk) \\ \Rightarrow && (g-k)^2 &> 3(h-gk) \end{align*} Pick \(g = h = 1\) and \(k = 1000\) then \((-999)^2 > 0 > 3(1-1000)\) so it is sufficient but not necessary.

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