The number \(\alpha\) is a common root of the
equations \(x^2 +ax +b=0\) and \(x^2+cx+d=0\)
(that is, \(\alpha\) satisfies both equations). Given that \(a\ne c\), show that
\[
\alpha =- \frac{b-d}{a-c}\,.
\]
Hence, or otherwise, show that the equations have at least one
common root if and only if
\[
(b-d)^2 -a(b-d)(a-c) + b(a-c)^2 =0\,.
\]
Does this result still hold if the condition \(a\ne c\) is not imposed?
Show that the equations
\(x^2+ax+b=0\) and \(x^3+(a+1)x^2+qx+r=0\)
have at least one common root if and only if
\[
(b-r)^2-a(b-r)(a+b-q) +b(a+b-q)^2=0\,.
\]
Hence, or otherwise, find the values of \(b\) for which the equations
\(2x^2+5 x+2 b=0\) and \(2x^3+7x^2+5x+1=0\)
have at least one common root.
Solution:
\begin{align*}
&& 0 &= \alpha^2 + a \alpha + b \tag{1} \\
&& 0 &= \alpha^2 + c \alpha + d \tag{2} \\
\\
(1) - (2): && 0 & =\alpha ( a-c) + (b-d) \\
\Rightarrow && \alpha &= - \frac{b-d}{a-c} \tag{\(a\neq c\)}
\end{align*}
(\(\Rightarrow\)) Suppose they have a common root, then given we know it's form, we must have:
\begin{align*}
&& 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\
\Rightarrow && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2
\end{align*}
(\(\Leftarrow\)) Suppose the equation holds, then
\begin{align*}
&& 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \\
\Rightarrow && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\
\end{align*}
So \(\alpha\) is a root of the first equation.
Considering \((1) - (2)\) we must have that \(\alpha(a-c) +(b-d) = t\) (whatever the second equation is), but that value is clearly \(0\), therefore \(\alpha\) is a root of both equations.
If \(a = c\) then the equation becomes \(0 = (b-d)^2\), ie the two equations are the same, therefore they must have common roots!
\begin{align*}
&& 0 &= x^2+ax+b \tag{1} \\
&& 0 &= x^3+(a+1)x^2+qx+r \tag{2} \\
\\
(2) - x(1) && 0 &= x^2 + (q-b)x + r \tag{3}
\end{align*}
Therefore if the equations have a common root, \((1)\) and \((3)\) have a common root, ie \((b-r)^2-a(b-r)(a-(q-b))+b(a-(q-b))^2 = 0\) which is exactly our condition.
\(a = \frac52, q = \frac52, r = \frac12\)
\begin{align*}
&& 0 &= \left (b-\frac12 \right)^2 - \frac52\left (b-\frac12\right) b + b^3 \\
&&&= b^2 -b + \frac14 - \frac52 b^2+\frac54b + b^3 \\
&&&= b^3 -\frac32 b^2 +\frac14 b + \frac14 \\\Rightarrow && 0 &= 4b^3 - 6b^2+b + 1 \\
&&&= (b-1)(4b^2-2b-1) \\
\Rightarrow && b &= 1, \frac{1 \pm \sqrt{5}}{4}\end{align*}