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2018 Paper 2 Q1
D: 1600.0 B: 1516.0
roots of polynomials quartic symmetric functions palindromic polynomial substitution reciprocal roots necessary and sufficient conditions

Show that, if \(k\) is a root of the quartic equation \[ x^4 + ax^3 + bx^2 + ax + 1 = 0\,, \tag{\(*\)} \] then \(k^{-1}\) is a root. You are now given that \(a\) and \(b\) in \((*)\) are both real and are such that the roots are all real.

  1. Write down all the values of \(a\) and \(b\) for which \((*)\) has only one distinct root.
  2. Given that \((*)\) has exactly three distinct roots, show that either \(b=2a-2\) or \(b=-2a-2\,\).
  3. Solve \((*)\) in the case \(b= 2 a -2\,\), giving your solutions in terms of \(a\).
Given that \(a\) and \(b\) are both real and that the roots of \((*)\) are all real, find necessary and sufficient conditions, in terms of \(a\) and \(b\), for \((*)\) to have exactly three distinct real roots.

Solution: Let \(f(x) = x^4 + ax^3 + bx^2 + ax + 1\), and suppose \(f(k) = 0\). Since \(f(0) = 1\), \(k \neq 0\), therefore we can talk about \(k^{-1}\). \begin{align*} && f(k^{-1}) &= k^{-4} + ak^{-3} + bk^{-2} + ak^{-1} + 1 \\ &&&= k^{-4}(1 + ak + bk^2 + ak^3 + k^4) \\ &&&= k^{-4}(k^4+ak^3+bk^2+ak+1) \\ &&&= k^{-4}f(k) = 0 \end{align*} Therefore \(k^{-1}\) is also a root of \(f\)

  1. If \(f\) has only on distinct root, we must have \(f(x) = (x+k)^4\) therefore \(k = k^{-1} \Rightarrow k^2 = 1 \Rightarrow k = \pm1\), or \(a = 4, b = 6\) or \(a = -4, b = 6\)
  2. If \(f\) has exactly three distinct roots then one of the roots must be a repeated \(1\) or \(-1\), ie \(0 = f(1) = 1 + a + b + a + 1 = 2 + b +2a \Rightarrow b = -2a-2\) or \(0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2\)
  3. If \(b = 2a-2\), we have \begin{align*} && f(x) &= 1 + ax + (2a-2)x^2 + ax^3 + x^4 \\ &&&= (x^2+2x+1)(1+(a-2)x+x^2) \\ \Rightarrow && x &= \frac{2-a \pm \sqrt{(a-2)^2 - 4}}{2} \\ &&&= \frac{2-a \pm \sqrt{a^2-4a}}{2} \end{align*}
\(f\) has exactly three distinct real roots iff \(b = \pm 2a - 2\) and \(b \neq 6\)

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1988 Paper 2 Q3
D: 1600.0 B: 1530.2
complex numbers roots of polynomials quadratic cubic symmetric functions reciprocal roots constraint analysis

The quadratic equation \(x^{2}+bx+c=0\), where \(b\) and \(c\) are real, has the properly that if \(k\) is a (possibly complex) root, then \(k^{-1}\) is a root. Determine carefully the restriction that this property places on \(b\) and \(c\). If, in addition to this property, the equation has the further property that if \(k\) is a root, then \(1-k\) is a root, find \(b\) and \(c\). Show that \[ x^{3}-\tfrac{3}{2}x^{2}-\tfrac{3}{2}x+1=0 \] is the only cubic equation of the form \(x^{3}+px^{2}+qx+r=0\), where \(p,q\) and \(r\) are real, which has both the above properties.

Solution: Suppose \(k\) is a root of our quadratic. There are two possibilities, if \(k^{-1} = k\) then we must have \(k^2 = 1\) so either \(\pm 1\) is a root or we must have \((x-k)(x-k^{-1}) = x^2+bx+c\). In the first case we can have: \begin{align*} x^2+bx +c = (x-1)^2 &\Rightarrow b = -2, c = 1 \\ x^2+bx +c = (x+1)^2 &\Rightarrow b = 2, c = 1 \\ x^2+bx +c = (x-1)(x+1) &\Rightarrow b = 0, c = 1 \\ \end{align*} In the other cases, \(c = 1\) and \(b = k^{-1}+k\). Therefore we must have \(c = 1\) and \(b\) can take any values. The statement "if \(k\) is a root then \(1-k\) is a root" implies these two roots are different, so we must have \(1-k = k^{-1} \Rightarrow k-k^2 = 1 \Rightarrow k^2-k+1 = 0\) so \(b = -1, c = 1\). Suppose \(x^3+px^2+qx+r = 0\) has the first property, then for any root \(k\) we must have: \(k^3 + pk^2 + qk + r = 0\) and \(1 + pk^{-1} + qk^{-2} + rk^{-3} = 0\) therefore \(x^3+px^2+qx+r\) and \(rx^3+qx^2+px+1 = 0\) must have identical roots (since \(x = \pm\) either wont work here since they imply having the roots \(1, 0\) or \(-1, 2, \frac12\) which is exactly our equation. Therefore \(r = 1, p = q\). Suppose \(x^3 + px^2 + px+1 = 0\) has the property that if \(k\) is a root \(1-k\) is a root, therefore: \begin{align*} 0 &= (1-k)^3+p(1-k)^2 + p(1-k) + 1 \\ &= 1 -3k+3k^2-k^3+p-2kp+pk^2+p-pk+1 \\ &= -k^3+(3+p)k^2+(-3-3p)k+(2+2p) \end{align*} Since these roots must be the same as the original roots, we must have \(3+p = -p, -3-3p = -p, 2+2p = -1 \Rightarrow p = -\frac32\)

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