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.
Write down all the values of \(a\) and \(b\) for which \((*)\) has only one distinct root.
Given that \((*)\) has exactly three distinct roots, show that either \(b=2a-2\) or \(b=-2a-2\,\).
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 \(f\)
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 \(a = -4, b = 6\)
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 \(0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2\)
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 \(b \neq 6\)