WAEC 2009 · Paper 2 · Q1

(a)

1. Write $2^{2x + 2} = 2^2 \times (2^x)^2$
   $= 4(2^x)^2$.
2. Let $y = 2^x$.
3. The equation becomes $4y^2 - 9y + 2 = 0$.
4. Factorise: $(4y - 1)(y - 2) = 0$, so $y = \frac14$ or $y = 2$.
5. $2^x = 2 = 2^1$ gives $x = 1$.
6. $2^x = \frac14 = 2^{-2}$ gives $x = -2$.
7. So $x = 1$ or $x = -2$.