Review question

# Does $(x-1)(x-2)\times \cdots \times (x-n) = k$ have a solution? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7017

## Solution

Given a positive integer $n$ and a real number $k$, consider the following equation in $x$, $(x-1)(x-2)(x-3) \times \cdots \times (x-n) = k.$ Which of the following statements about this equation is true?

1. If $n=3$, then the equation has no real solution $x$ for some values of $k$.

If $n=3$, we have the equation $(x-1)(x-2)(x-3) = k$, which is a cubic function in $x$.

We know from the shape of cubics that the function will take on all real values, and so (a) is untrue.

Useful fact: every cubic equation with real coefficients has at least one real root.

1. If $n$ is even, then the equation has a real solution $x$ for any given value of $k$.

If $n$ is even, then the leading term of the function will be a positive even power of $x$, such as a quadratic ($n=2$) or a quartic ($n=4$) function.

We know from the shape of these functions that they have a finite minimum value.

Thus the equation will not have a solution for sufficiently negative values of $k$.

Hence (b) is untrue.

1. The equation never has a repeated solution $x$ for any given values of $k$ and $n$.

If we set $n=2$, we have $(x-1)(x-2) = k$, i.e. $x^2 - 3x + (2-k)=0$. Completing the square gives $\left(x-\frac{3}{2}\right)^2 -\left(\frac{1}{4} + k\right) = 0,$ so if we set $k = -\dfrac{1}{4}$, then the equation has a repeated root at $x = \dfrac{3}{2}$.

Another way of thinking about this is that when $n=2$, the graph $y=(x-1)(x-2)$ is a parabola, so it has a minimum point.

When $y=k$ touches the graph at this minimum point, there will be a repeated root, and so this curve is a counter-example to (d).

Hence (d) is untrue, and by process of elimination, the answer is (c).

We can also show that (c) is true directly.

1. If $k \ge 0$ then the equation has (at least) one real solution $x$.

If we let $f(x) = (x-1)(x-2)(x-3) \times \cdots \times (x-n)$ then $f(n) = 0$.

If we then increase $x$ from $n$, $f(x)$ also increases from $0$, since every term in the product is positive and increasing.

Hence $f(x)$ will take every positive real value for $x>n$, and so the equation will have some solution for every $k\ge0$, so (c) is true.

This is also clear from the shape of the graphs of $y=f(x)$.