WAEC 2022 · Paper 2 · Q2

(a)
Solve $2^{(2y + 1)} - 5(2^y) + 2 = 0$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Split the index: $2^{2y + 1} = 2 \times (2^y)^2$ .
Let $x = 2^y$ : $2x^2 - 5x + 2 = 0$ .
Factorise: $(2x - 1)(x - 2) = 0$ , so $x = \frac12$ or $x = 2$ .
Back to $y$ : $2^y = \frac12 = 2^{-1}$ gives $y = -1$ , and $2^y = 2$ gives $y = 1$ .