Prove that, if \(a\) is real, the equation \[\begin{equation} (x^2+1)(x+1)=(a^2+1)(a+1) \label{eq:1} \end{equation}\]

has one and only one real root.

It is easy to see that \(x=a\) is a root of the equation so we must show that there are no other solutions.

Since \(x=a\) is a root, we must be able to rewrite \(\eqref{eq:1}\) in the form \((x-a)\:f(x)=0\), where \(f(x)\) is a quadratic polynomial.

We’ll then be able to check whether \(f(x)=0\) has any roots, and therefore whether \(\eqref{eq:1}\) has any other roots besides \(x=a\).

We can expand \(\eqref{eq:1}\) as \[\begin{align*} x^3+x^2+&x+1 = a^3+a^2+a+1\\ x^3+x^2+&x -a(a^2+a+1) =0. \end{align*}\] To factorise this we can easily write down the \(x^2\) and constant terms of \(f(x)\). We can then find the \(x\) term by comparing coefficients of \(x^2\). \[\begin{align*} x^3+x^2+x -a(a^2+a+1) &= (x-a)\left(x^2 + ...x+(a^2+a+1)\right)\\ &= (x-a)\left(x^2 + (a+1)x+(a^2+a+1)\right) \end{align*}\] We now aim to show that \[\begin{equation} x^2+(a+1)x+(a^2+a+1)=0 \label{eq:2} \end{equation}\]

has no real roots.

We can calculate the discriminant of \(\eqref{eq:2}\) to be \[ \Delta = (a+1)^2-4(a^2+a+1)=a^2+2a+1-4a^2-4a-4=-3a^2-2a-3. \] Completing the square, this becomes \[\Delta = -3\left(a+\frac{1}{3}\right)^2-\frac{8}{3},\] which is negative for all real \(a\).

Therefore \(\eqref{eq:2}\) has no real roots and \(\eqref{eq:1}\) has one, and only one real root, which is \(x=a\).

Alternative solution using calculus

Multiplying out the brackets on the left hand side of \(\eqref{eq:1}\) we have \[(x^2+1)(x+1)=x^3+x^2+x+1\] and the right hand side is a constant.

Consider the graph of \[y=x^3+x^2+x.\]

We can find its turning points by differentiating. \[\frac{dy}{dx}=3x^2+2x+1\] This quadratic expression has a negative discriminant and therefore no real zeros. So we conclude that the graph of the function has no stationary points and therefore cuts the \(x\)-axis exactly once.

The right hand side of \(\eqref{eq:1}\) has the effect of translating the graph vertically and does not change the number of axis intercepts. So we conclude that the equation \(\eqref{eq:1}\) has one and only one real root.

This applet shows the graph of \(y = (x^2+1)(x+1)-(a^2+1)(a+1)\).