IMC 2018 Q21
21. For certain values of \(x\), the list \(x\), \(x + 6\) and \(x^2\) contains just two different numbers.
How many such values of \(x\) are there?
- A 1
- B 2
- C 3
- D 4
- E 5
Intermediate Mathematical Challenge 2018, Question 21. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D 4
Clearly, there are no values of \(x\) for which \(x = x + 6\). So the possible values of \(x\) will be solutions of the equation \(x = x^2\) or the equation \(x + 6 = x^2\).
Consider the first equation: \(x = x^2\). This simplifies to \(x(1 - x) = 0\). So \(x = 0\) or \(x = 1\).
The equation \(x + 6 = x^2\) simplifies to \(x^2 - x - 6 = 0\). So \((x - 3)(x + 2) = 0\).
Therefore \(x = 3\) or \(x = -2\). So there are four different values of \(x\).
Solution © UK Mathematics Trust, from the IMC 2018 solutions. Longer versions with investigations: Solutions and investigations (PDF).