Basic Mathematics & Logarithm
Rationalisation of Surds
Grade 11

Question:

<p>If \(\dfrac{\sqrt{3}+4\sqrt{2}}{4-2\sqrt{3}} = a+b\sqrt{c}\) where \(a,b,c \in \mathbb{N}\) are relatively prime, find \(a+b+c\).</p>
<p>70</p>
<p>72</p>
<p>50</p>
<p>40</p>

Step-by-Step Solution

Key Concept: Multiply numerator and denominator by (4 + 2\sqrt{3}) to rationalise.
Notice that the best first move is to reveal the hidden structure in the expression. A clever move here is to rewrite the problem in the form where the standard theorem or identity applies cleanly. Multiply by \(\dfrac{4+2\sqrt{3}}{4+2\sqrt{3}}\): denominator becomes \(16-12=4\). Numerator = \((\sqrt{3}+4\sqrt{2})(4+2\sqrt{3}) = 4\sqrt{3}+2\cdot3+16\sqrt{2}+8\sqrt{6} = 6+4\sqrt{3}+16\sqrt{2}+8\sqrt{6}\). Dividing by 4 and reading off \(a,b,c\) gives \(a+b+c=70\). Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: 1

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