NECO 2022 · Paper 2 · Q1

(a)
If $\sqrt{3} = 1.73$ , rationalise $\frac{5}{\sqrt{3}}$ correct to 2 significant figures.
(b)
Solve the quadratic equation $3x^2 - 20x + 12 = 0$ , using the completing the square method.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Multiply the top and bottom by $\sqrt3$ to clear the surd from the denominator: $\frac{5}{\sqrt3} = \frac{5\sqrt3}{3}$ .
Put in $\sqrt3 = 1.73$ : the numerator is $5 \times 1.73 = 8.65$ .
Divide by 3: $8.65 \div 3 = 2.883\ldots$ , which is $2.9$ to 2 significant figures.

Divide every term by 3 so that $x^2$ has coefficient 1: $x^2 - \frac{20}{3}x + 4 = 0$ .
Move the constant to the right: $x^2 - \frac{20}{3}x = -4$ .
Add the square of half the coefficient of $x$ , $\left(\frac{10}{3}\right)^2 = \frac{100}{9}$ , to both sides: $x^2 - \frac{20}{3}x + \frac{100}{9} = \frac{100}{9} - 4$ .
Write the left side as a square and simplify the right: $\left(x - \frac{10}{3}\right)^2 = \frac{64}{9}$ .
Take square roots, keeping both signs: $x - \frac{10}{3} = \pm\frac{8}{3}$ .
So $x = \frac{10}{3} + \frac83 = 6$ or $x = \frac{10}{3} - \frac83 = \frac23$ .