WASSCE Surds is one of the topics in WASSCE mathematics syllabus. The followings are WASSCE past questions and solutions on Surds. Candidates are advised to try each question by themselves before viewing solution. Frequent practice guarantees success in exams.MATHS LIBRARY FOR MORE TOPICS
WASSCE 2014 school candidates. Simplify \(3\sqrt{75}-\sqrt{12}+\sqrt{108},\) leaving the answer in surd form (radicals).
Solution Solution:
\(3\sqrt{75}-\sqrt{12}+\sqrt{108}\)
\(=3(\sqrt{25 \times 3}-\sqrt{4 \times 3}+\sqrt{36 \times 3})\)
\(=3(5\sqrt{3})-2\sqrt{3}+6\sqrt{3}\)
\(=15\sqrt{3}-2\sqrt{3}+6\sqrt{3}\)
\(=19\sqrt{3}\)
By: Abdul Karim
WASSCE 2013 school candidates. Simplify \(\frac{3}{4}\sqrt{128}-\sqrt{50}\) leaving your answer in surd form.
Solution Solution:
\(\frac{3}{4}\sqrt{128}-\sqrt{50}\)
\(=\frac{3}{4}(\sqrt{64 \times 2})-\sqrt{25 \times 2}\)
\(=\frac{3}{4}(8\sqrt{2})-5\sqrt{2}\)
\(=6\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\)
By: Abdul Karim
WASSCE 2011 school candidates. Simplify \(\frac{2}{3+2\sqrt{2}}+\frac{1}{3-2\sqrt{2}}\) leaving your answer in the form \(a+b\sqrt{2}.\)
Solution Solution:
\(\frac{2}{3+2\sqrt{2}}+\frac{1}{3-2\sqrt{2}}\)
\(=\frac{2(3-2\sqrt{2})+1(3+2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}\) Hint Find LCM \(\Rightarrow (3-2\sqrt{2})(3-2\sqrt{2})\)
\(=\frac{6-4\sqrt{2}+3+2\sqrt{2}}{9-(2\sqrt{2})^2}\)
\(=\frac{-4\sqrt{2}+2\sqrt{2}+6+3}{9-(4\sqrt{2})}\)
\(=\frac{-2\sqrt{2}+9}{9-8}\)
\(=\frac{9-2\sqrt{2}}{1}\)
\(=9-2\sqrt{2} \) in the form of \( a+b\sqrt{2} \) where \( a=9 \) and \( b=-2.\)
By: Abdul Karim
WASSCE 2008 school candidates. Simplify: \((\sqrt{5})^{-2}\times75^{\frac{1}{2}}\times12^{-\frac{1}{2}}.\)
Solution Solution:
\((\sqrt{5})^{-2}\times75^{\frac{1}{2}}\times12^{-\frac{1}{2}}\) Hint Note: Power \(\frac{1}{2}\) means square root.
\(\Rightarrow (\frac{1}{\sqrt{5}})^2 \times \sqrt{75} \times \frac{1}{\sqrt{12}}\)
\(=\frac{1}{5} \times \sqrt{25 \times 3} \times \frac{1}{\sqrt{4 \times 3}}\)
\(=\frac{1}{5} \times 5\sqrt{3} \times \frac{1}{2\sqrt{3}}\)
\(=\sqrt{3} \times \frac{1}{2}\sqrt{3}\) Hint \(5 \text{ cancelled } 5\)
\(=\frac{1}{2}.\) Hint \(\sqrt{3} \text{ cancelled } \sqrt{3}\)
By: Abdul Karim
WASSCE 2007 private candidates. Without using four-figure tables or calculator, simplify: \(\sqrt{50}-3\sqrt{2}(2\sqrt{2}-5)-5\sqrt{32}.\)
Solution Solution:
\(\sqrt{50}-3\sqrt{2}(2\sqrt{2}-5)-5\sqrt{32}\)
\(=\sqrt{25 \times 2}\)\(-(3 \times 2)(\sqrt{2})^2\)\(+(3 \times 5)\sqrt{2}\)\(-5(\sqrt{16 \times 2})\)
\(=5\sqrt{2}-6(2)+15\sqrt{2}-5(4\sqrt{2})\)
\(=5\sqrt{2}-12+15\sqrt{2}-20\sqrt{2}\)
\(=5\sqrt{2}+15\sqrt{2}-20\sqrt{2}-12\)
\(=(5+15-20)\sqrt{2}-12\)
\(=0\sqrt{2}-12\)
\(=-12.\)
By: Abdul Karim
WASSCE 2006 private candidates. Simplify: \((14+3\sqrt{7})(2-\frac{3}{\sqrt{7}}).\)
Solution Solution:
\((14+3\sqrt{7})(2-\frac{3}{\sqrt{7}})\)
\(=14(2-\frac{3}{\sqrt{7}})+3\sqrt{7}(2-\frac{3}{\sqrt{7}})\)
\(=28-\frac{42}{\sqrt{7}}+6\sqrt{7}-\frac{3 \times 3\sqrt{7}}{\sqrt{7}}\)
\(=28-\frac{42}{\sqrt{7}}+6\sqrt{7}-9\)
\(=6\sqrt{7}\frac{42}{\sqrt{7}}+28-9\)
\(=\frac{6(\sqrt{7})^2-42}{\sqrt{7}}+19\)
\(=\frac{6(7)-42}{\sqrt{7}}+19\)
\(=\frac{42-42}{\sqrt{7}}+19\)
\(=\frac{0}{\sqrt{7}}+19\)
\(=19.\)
By: Abdul Karim
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X^1/3+x^1/2=12 find x
@muftifaddal, @saanimc