WAEC 2018 · Paper 2 · Q4

(a)
If $\sqrt{3x + 1} - \sqrt{2x - 1} = 1$ , find the values of $x$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Get one root on its own: $\sqrt{3x + 1} = 1 + \sqrt{2x - 1}$ .
Square both sides, keeping the middle term on the right: $3x + 1 = 1 + 2\sqrt{2x - 1} + 2x - 1$ .
Simplify and get the root on its own: $x + 1 = 2\sqrt{2x - 1}$ .
Square again: $x^2 + 2x + 1 = 4(2x - 1)$ , so $x^2 - 6x + 5 = 0$ .
Factorise: $(x - 1)(x - 5) = 0$ , so $x = 1$ or $x = 5$ .
Check in the original equation: $x = 1$ gives $2 - 1 = 1$ ✓ and $x = 5$ gives $4 - 3 = 1$ ✓.
So $x = 1$ or $x = 5$ .