Review question

# Can we show $(x^2+1)(x+1)=(a^2+1)(a+1)$ has just one real root? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9108

## Solution

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

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 $$$x^2+(a+1)x+(a^2+a+1)=0 \label{eq:2}$$$

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 $y = (x^2+1)(x+1)-(a^2+1)(a+1)$.