WAEC Further Mathematics 2014 Paper

Concept:Rationalizing the denominator removes the surd from the bottom of the fraction by using the conjugate.Explanation:First, simplify $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.So the expression becomes $\frac{1 + 2\sqrt{2}}{3 - \sqrt{2}}$.Multiply the numerator and denominator by the conjugate $3 + \sqrt{2}$.This gives $\frac{1 + 2\sqrt{2}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}$.Expand the numerator: $1 \cdot 3 + 1 \cdot \sqrt{2} + 2\sqrt{2} \cdot 3 + 2\sqrt{2} \cdot \sqrt{2}$.So the numerator equals $3 + \sqrt{2} + 6\sqrt{2} + 4 = 7 + 7\sqrt{2}$.The denominator is $(3 - \sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7$.Therefore the fraction simplifies to $\frac{7 + 7\sqrt{2}}{7} = 1 + \sqrt{2}$.Answer:$1 + \sqrt{2}$ (Option D).