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1999 Paper 2 Q2
D: 1600.0 B: 1516.0
quadratics discriminant inequality no real roots proof parameter analysis

Consider the quadratic equation $$ nx^2+2x \sqrt{pn^2+q} + rn + s = 0, \tag{*} $$ where \(p>0\), \(p\neq r\) and \(n=1\), \(2\), \(3\), \(\ldots\) .

  1. For the case where \(p=3\), \(q=50\), \(r=2\), \(s=15\), find the set of values of \(n\) for which equation \((*)\) has no real roots.
  2. Prove that if \(p < r\) and \(4q(p-r) > s^2\), then \((*)\) has no real roots for any value of \(n\).
  3. If \(n=1\), \(p-r=1\) and \(q={s^2}/8\), show that \((*)\) has real roots if, and only if, \(s \le 4-2\sqrt{2}\ \) or \(\ s \ge 4+2\sqrt{2}\).

Solution:

  1. \(\,\) \begin{align*} && 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\ && 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\ \Leftrightarrow && 0 &> n^2-15n+5\\ \text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\ &&&\approx \frac{15\pm14.x}{2}\\ \Leftrightarrow &&n &\in \{1, 2, \cdots, 14\} \end{align*}
  2. \(\,\) \begin{align*} && 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\ \Leftrightarrow && 0&>(p-r)n^2-sn+q \end{align*} Which is always true if \(r > p\) and \(s^2 < 4q(p-r)\)
  3. \(\,\) \begin{align*} && 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\ && 0 &\leq \Delta = 4(p+q) - 4(r+s) \\ && 0 &\leq 1 + s^2/8 - s \\ \text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\ &&&= 4 \pm 4\sqrt{\frac12} \\ &&&= 4 \pm 2\sqrt{2} \\ \Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2} \end{align*}

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1990 Paper 2 Q1
D: 1600.0 B: 1516.0
inequality quadratics quartic completing the square proof no real roots case analysis polynomials

Prove that both \(x^{4}-2x^{3}+x^{2}\) and \(x^{2}-8x+17\) are non-negative for all real \(x\). By considering the intervals \(x\leqslant0\), \(0 < x\leqslant2\) and \(x > 2\) separately, or otherwise, prove that the equation \[ x^{4}-2x^{3}+x^{2}-8x+17=0 \] has no real roots. Prove that the equation \(x^{4}-x^{3}+x^{2}-4x+4=0\) has no real roots.

Solution: \begin{align*} x^4 - 2x^3+x^2 &= x^2(x^2-2x+1) \\ &= x^2(x-1)^2 > 0 \end{align*} Since \(x\)and \(x-1\) can't both be zero, and square cannot be negative. \begin{align*} x^2 - 8x+17 &= (x-4)^2 +1 \geq 1 > 0 \end{align*} If \(x \leq 2\) then \(x^4 - 2x^3+x^2 > 0\) and \(17-8x \geq 1\) so \(x^4-2x^3+x^2-8x+17 > 0\) If \(x > 2\) then \(x^4-2x^3 = x^3(x-2) \geq 0\) and \(x^2-8x+17 > 0\) so \(x^4-2x^3+x^2-8x+17 > 0\), so for all real numbers our polynomial is positive and therefore cannot have any roots. Note that: \(x^4-x^3+x^2 = x^2(x^2-x+1) > 0\) and \(x^2-4x+4 =(x-2)^2 \geq 0\) If \(x < 1\) then \(x^4-x^3+x^2 > 0\) and \(4(1-x) > 0\) so \(x^4-x^3+x^2-4x+4 > 0\). If \(x > 1\) then \(x^4-x^3 = x^3(x-1) > 0\) and \(x^2-4x+4 \geq 0\) therefore \(x^4-x^3+x^2-4x+4 > 0\). Therefore \(x^4-x^3+x^2-4x+4 > 0\) for all real \(x\) and hence there are no real roots.

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