WAEC 2011 · Paper 2 · Q4

(a)
Solve $2^{(2y + 2)} - 9(2^y) = -2$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Split the index: $2^{2y + 2} = 2^2 \times 2^{2y} = 4(2^y)^2$ .
So the equation is $4(2^y)^2 - 9(2^y) = -2$ .
Let $x = 2^y$ : $4x^2 - 9x = -2$ .
Add 2 to both sides: $4x^2 - 9x + 2 = 0$ .
Factorise: $(4x - 1)(x - 2) = 0$ , so $x = \frac14$ or $x = 2$ .
Back to $y$ : $2^y = \frac14 = 2^{-2}$ gives $y = -2$ , and $2^y = 2 = 2^1$ gives $y = 1$ .