4.2 Practice Problems
- (a).
- \(2\sqrt {27}-3\sqrt {48}+\sqrt {75}\)
- (b).
- \(\sqrt {80}-\sqrt {20}-2\sqrt {45}\)
- (c).
- \(\frac {\sqrt {28}}{\sqrt {175}}\)
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Solution. The method is the same throughout: pull the largest square factor out of each root so that every term carries the same surd, then collect.
(a). \begin {align*} 2\sqrt {27}-3\sqrt {48}+\sqrt {75} &= 2\sqrt {9\times 3}-3\sqrt {16\times 3}+\sqrt {25\times 3}\\ &= 2(3)\sqrt {3}-3(4)\sqrt {3}+5\sqrt {3}\\ &= 6\sqrt {3}-12\sqrt {3}+5\sqrt {3}=-\sqrt {3}. \end {align*}
(b). \begin {align*} \sqrt {80}-\sqrt {20}-2\sqrt {45} &= \sqrt {16\times 5}-\sqrt {4\times 5}-2\sqrt {9\times 5}\\ &= 4\sqrt {5}-2\sqrt {5}-6\sqrt {5}=-4\sqrt {5}. \end {align*}
(c). \(\frac {\sqrt {28}}{\sqrt {175}}=\sqrt {\frac {28}{175}} =\sqrt {\frac {4}{25}}=\frac {2}{5}\), cancelling the common factor \(7\) inside the root before taking it.
Note 4.7. Two habits are worth forming here. In (a) and (b), the terms can only be collected once they all carry the same surd — \(\sqrt {80}\) and \(\sqrt {20}\) cannot be added as they stand, but \(4\sqrt {5}\) and \(2\sqrt {5}\) can. And in (c), combining the two roots into one before simplifying turns an awkward division into a cancellation.
Check (b) numerically if the collecting step feels uncertain: \(\sqrt {80}\approx 8.944\), \(\sqrt {20}\approx 4.472\), \(2\sqrt {45}\approx 13.416\), and \(8.944-4.472-13.416=-8.944\), which is \(-4\sqrt {5}\). A ten-second check of this kind catches a slip in the final subtraction, which is where they happen.
Problem 4.2. Rationalise the denominator:
- (a).
- \(\frac {\sqrt {2}-1}{1+\sqrt {2}}\)
- (b).
- \(\frac {\sqrt {3}-2}{\sqrt {3}-1}\)
- (c).
- \(\frac {2+\sqrt {2}}{\left (\sqrt {2}-1\right )^{2}}\)
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Solution. (a). Multiply above and below by the conjugate \(1-\sqrt {2}\): \[\frac {\left (\sqrt {2}-1\right )\left (1-\sqrt {2}\right )} {\left (1+\sqrt {2}\right )\left (1-\sqrt {2}\right )} =\frac {\sqrt {2}-2-1+\sqrt {2}}{1-2} =\frac {2\sqrt {2}-3}{-1}=3-2\sqrt {2}.\]
(b). Conjugate of the denominator is \(\sqrt {3}+1\): \[\frac {\left (\sqrt {3}-2\right )\left (\sqrt {3}+1\right )} {\left (\sqrt {3}-1\right )\left (\sqrt {3}+1\right )} =\frac {3+\sqrt {3}-2\sqrt {3}-2}{3-1} =\frac {1-\sqrt {3}}{2}.\]
(c). Expand the denominator first: \(\left (\sqrt {2}-1\right )^{2}=2-2\sqrt {2}+1=3-2\sqrt {2}\). Now multiply above and below by \(3+2\sqrt {2}\): \[\frac {\left (2+\sqrt {2}\right )\left (3+2\sqrt {2}\right )}{\left (3-2\sqrt {2}\right )\left (3+2\sqrt {2}\right )} =\frac {6+4\sqrt {2}+3\sqrt {2}+4}{9-8} =10+7\sqrt {2}.\]
Problem 4.3. Simplify, giving your answer in the form \(p+q\sqrt {3}\) with \(p,q\in \mathbb {R}\): \[\frac {3}{2-\sqrt {3}}+\sqrt {27}-5\sqrt {48}.\]
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Solution. Deal with the three terms separately. Rationalising the first, \[\frac {3}{2-\sqrt {3}}=\frac {3\left (2+\sqrt {3}\right )}{\left (2-\sqrt {3}\right )\left (2+\sqrt {3}\right )} =\frac {6+3\sqrt {3}}{4-3}=6+3\sqrt {3}.\] For the others, \(\sqrt {27}=\sqrt {9\times 3}=3\sqrt {3}\) and \(5\sqrt {48}=5\sqrt {16\times 3}=20\sqrt {3}\). Hence \[\frac {3}{2-\sqrt {3}}+\sqrt {27}-5\sqrt {48} =6+3\sqrt {3}+3\sqrt {3}-20\sqrt {3}=6-14\sqrt {3},\] so \(p=6\) and \(q=-14\).
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