PRELIMINARY CONTEST 2024
Contest 10
Adonten SHS – 54 pts (Winner)
Yaa Asantewaa Girls’ SHS – 36 pts
Tolon SHS – 9 pts
Contest 11
Amaniampong SHS – 55 pts (Winner)
New Juaben SHS – 39 pts
St. Mary’s SHS – 22 pts
Contest 12
Suhum Presby SHS – 59 pts (Winner)
St. Charles Seminary SHS – 28 pts
Fiaseman SHS – 21 pts
ROUND 1
PREAMBLE
Simplify the given expressions containing surds.
FORMULA
$\dfrac{\sqrt{a}\pm\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}=\dfrac{a+b\pm2\sqrt{ab}}{a-b}$
1. Simplify $\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}$
ANSWER: $\dfrac{5+\sqrt{21}}{2}$
SOLUTION 1
Given $a=7$ and $b=3$:
$\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}=\dfrac{7+3+2\sqrt{7\times3}}{7-3}$
$=\dfrac{10+2\sqrt{21}}{4}$
$=\dfrac{2(5+\sqrt{21})}{4}$
$=\dfrac{5+\sqrt{21}}{2}$
SOLUTION 2
Multiply top and bottom by the conjugate of the denominator, $\sqrt{7}+\sqrt{3}$:
$\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}\times\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}$
$=\dfrac{(\sqrt{7})^2+2(\sqrt{7})(\sqrt{3})+(\sqrt{3})^2}{(\sqrt{7})^2-(\sqrt{3})^2}$
$=\dfrac{7+3+2\sqrt{21}}{7-3}$
$=\dfrac{10+2\sqrt{21}}{4}$
$=\dfrac{5+\sqrt{21}}{2}$
2. Simplify $\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}$
ANSWER: $\dfrac{9-\sqrt{77}}{2}$
SOLUTION 1
Given $a=11$ and $b=7$:
$\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=\dfrac{11+7-2\sqrt{11\times7}}{11-7}$
$=\dfrac{18-2\sqrt{77}}{4}$
$=\dfrac{2(9-\sqrt{77})}{4}$
$=\dfrac{9-\sqrt{77}}{2}$
SOLUTION 2
Multiply top and bottom by $\sqrt{11}-\sqrt{7}$:
$\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}\times\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}-\sqrt{7}}$
$=\dfrac{(\sqrt{11})^2-2(\sqrt{11})(\sqrt{7})+(\sqrt{7})^2}{(\sqrt{11})^2-(\sqrt{7})^2}$
$=\dfrac{11+7-2\sqrt{77}}{11-7}$
$=\dfrac{18-2\sqrt{77}}{4}$
$=\dfrac{9-\sqrt{77}}{2}$
3. Simplify $\dfrac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}$
ANSWER: $12-\sqrt{143}$
SOLUTION 1
Given $a=13$ and $b=11$:
$\dfrac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}=\dfrac{13+11-2\sqrt{13\times11}}{13-11}$
$=\dfrac{24-2\sqrt{143}}{2}$
$=12-\sqrt{143}$
SOLUTION 2
Multiply top and bottom by $\sqrt{13}-\sqrt{11}$:
$\dfrac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}\times\dfrac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}$
$=\dfrac{(\sqrt{13})^2-2(\sqrt{13})(\sqrt{11})+(\sqrt{11})^2}{(\sqrt{13})^2-(\sqrt{11})^2}$
$=\dfrac{13+11-2\sqrt{143}}{13-11}$
$=\dfrac{24-2\sqrt{143}}{2}$
$=12-\sqrt{143}$
PRACTICE QUESTIONS
1. Simplify $\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$.
ANSWER: $\dfrac{7+2\sqrt{10}}{3}$
SOLUTION 1
Use $\dfrac{\sqrt{a}\pm\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}=\dfrac{a+b\pm2\sqrt{ab}}{a-b}$ with $a=5$ and $b=2$:
$\dfrac{5+2+2\sqrt{5\times2}}{5-2}$
$=\dfrac{7+2\sqrt{10}}{3}$
$=\dfrac{7+2\sqrt{10}}{3}$
SOLUTION 2
Multiply top and bottom by the conjugate of the denominator, $\sqrt{5}+\sqrt{2}$:
$\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$
$=\dfrac{(\sqrt{5})^2+2\sqrt{5}\sqrt{2}+(\sqrt{2})^2}{(\sqrt{5})^2-(\sqrt{2})^2}$
$=\dfrac{7+2\sqrt{10}}{3}$
$=\dfrac{7+2\sqrt{10}}{3}$
2. Simplify $\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}$.
ANSWER: $\dfrac{9-\sqrt{77}}{2}$
SOLUTION 1
Use $\dfrac{\sqrt{a}\pm\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}=\dfrac{a+b\pm2\sqrt{ab}}{a-b}$ with $a=11$ and $b=7$:
$\dfrac{11+7-2\sqrt{11\times7}}{11-7}$
$=\dfrac{18-2\sqrt{77}}{4}$
$=\dfrac{9-\sqrt{77}}{2}$
SOLUTION 2
Multiply top and bottom by the conjugate of the denominator, $\sqrt{11}-\sqrt{7}$:
$\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}\times\dfrac{\sqrt{11}-\sqrt{7}}{\sqrt{11}-\sqrt{7}}$
$=\dfrac{(\sqrt{11})^2-2\sqrt{11}\sqrt{7}+(\sqrt{7})^2}{(\sqrt{11})^2-(\sqrt{7})^2}$
$=\dfrac{18-2\sqrt{77}}{4}$
$=\dfrac{9-\sqrt{77}}{2}$
3. Simplify $\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}$.
ANSWER: $2+\sqrt{3}$
SOLUTION 1
Use $\dfrac{\sqrt{a}\pm\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}=\dfrac{a+b\pm2\sqrt{ab}}{a-b}$ with $a=6$ and $b=2$:
$\dfrac{6+2+2\sqrt{6\times2}}{6-2}$
$=\dfrac{8+2\sqrt{12}}{4}$
$=2+\sqrt{3}$
SOLUTION 2
Multiply top and bottom by the conjugate of the denominator, $\sqrt{6}+\sqrt{2}$:
$\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\times\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}$
$=\dfrac{(\sqrt{6})^2+2\sqrt{6}\sqrt{2}+(\sqrt{2})^2}{(\sqrt{6})^2-(\sqrt{2})^2}$
$=\dfrac{8+2\sqrt{12}}{4}$
$=2+\sqrt{3}$
4. Simplify $\dfrac{\sqrt{13}-\sqrt{5}}{\sqrt{13}+\sqrt{5}}$.
ANSWER: $\dfrac{9-\sqrt{65}}{4}$
SOLUTION 1
Use $\dfrac{\sqrt{a}\pm\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}=\dfrac{a+b\pm2\sqrt{ab}}{a-b}$ with $a=13$ and $b=5$:
$\dfrac{13+5-2\sqrt{13\times5}}{13-5}$
$=\dfrac{18-2\sqrt{65}}{8}$
$=\dfrac{9-\sqrt{65}}{4}$
SOLUTION 2
Multiply top and bottom by the conjugate of the denominator, $\sqrt{13}-\sqrt{5}$:
$\dfrac{\sqrt{13}-\sqrt{5}}{\sqrt{13}+\sqrt{5}}\times\dfrac{\sqrt{13}-\sqrt{5}}{\sqrt{13}-\sqrt{5}}$
$=\dfrac{(\sqrt{13})^2-2\sqrt{13}\sqrt{5}+(\sqrt{5})^2}{(\sqrt{13})^2-(\sqrt{5})^2}$
$=\dfrac{18-2\sqrt{65}}{8}$
$=\dfrac{9-\sqrt{65}}{4}$
5. Simplify $\dfrac{\sqrt{10}-\sqrt{3}}{\sqrt{10}+\sqrt{3}}$.
ANSWER: $\dfrac{13-2\sqrt{30}}{7}$
SOLUTION 1
Use $\dfrac{\sqrt{a}\pm\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}=\dfrac{a+b\pm2\sqrt{ab}}{a-b}$ with $a=10$ and $b=3$:
$\dfrac{10+3-2\sqrt{10\times3}}{10-3}$
$=\dfrac{13-2\sqrt{30}}{7}$
$=\dfrac{13-2\sqrt{30}}{7}$
SOLUTION 2
Multiply top and bottom by the conjugate of the denominator, $\sqrt{10}-\sqrt{3}$:
$\dfrac{\sqrt{10}-\sqrt{3}}{\sqrt{10}+\sqrt{3}}\times\dfrac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}$
$=\dfrac{(\sqrt{10})^2-2\sqrt{10}\sqrt{3}+(\sqrt{3})^2}{(\sqrt{10})^2-(\sqrt{3})^2}$
$=\dfrac{13-2\sqrt{30}}{7}$
$=\dfrac{13-2\sqrt{30}}{7}$
6. Simplify $\dfrac{\sqrt{8}+\sqrt{5}}{\sqrt{8}-\sqrt{5}}$.
ANSWER: $\dfrac{13+4\sqrt{10}}{3}$
SOLUTION 1
Use $\dfrac{\sqrt{a}\pm\sqrt{b}}{\sqrt{a}\mp\sqrt{b}}=\dfrac{a+b\pm2\sqrt{ab}}{a-b}$ with $a=8$ and $b=5$:
$\dfrac{8+5+2\sqrt{8\times5}}{8-5}$
$=\dfrac{13+2\sqrt{40}}{3}$
$=\dfrac{13+4\sqrt{10}}{3}$
SOLUTION 2
Multiply top and bottom by the conjugate of the denominator, $\sqrt{8}+\sqrt{5}$:
$\dfrac{\sqrt{8}+\sqrt{5}}{\sqrt{8}-\sqrt{5}}\times\dfrac{\sqrt{8}+\sqrt{5}}{\sqrt{8}+\sqrt{5}}$
$=\dfrac{(\sqrt{8})^2+2\sqrt{8}\sqrt{5}+(\sqrt{5})^2}{(\sqrt{8})^2-(\sqrt{5})^2}$
$=\dfrac{13+2\sqrt{40}}{3}$
$=\dfrac{13+4\sqrt{10}}{3}$