WAEC 2014 · Paper 2 · Q6

(a)
The deviations from the mean of a set of numbers are $-2$ , $x$ , $(x + 7)$ , $(x + 2)^2$ and $(x + 3)^2$ , where $x$ is a constant. Find the value of $x$ .

Worked solution (try it first)

The deviations from the mean always add up to 0: $-2 + x + (x + 7) + (x + 2)^2 + (x + 3)^2 = 0$ .
Expand the squares: $(x + 2)^2 = x^2 + 4x + 4$ and $(x + 3)^2 = x^2 + 6x + 9$ .
Collect the $x^2$ terms: $2x^2$ .
The $x$ terms: $x + x + 4x + 6x = 12x$ .
The numbers: $-2 + 7 + 4 + 9 = 18$ .
So $2x^2 + 12x + 18 = 0$ .
Divide by 2: $x^2 + 6x + 9 = 0$ .
Factorise: $(x + 3)^2 = 0$ , so $x = -3$ .
Check: the deviations are $-2, -3, 4, 1, 0$ , which add up to 0 ✓.