PRELIMINARY STAGE
2026
Chemo Senior High Technical School: 50 points (Winner after tiebreaker, qualified to the one-eighth stage)
Xavier SHS: 50 points (Lost tiebreaker)
Wallace Academy: 16 points
All three schools won the NSMQ Star and the Clean Sheet Prize.
QUESTION
Rationalize the denominator of $\dfrac{5-2\sqrt3}{5+2\sqrt3}$.
ANSWER: $\dfrac{37-20\sqrt3}{13}$
SOLUTION 1
Multiply the numerator and denominator by the conjugate of the denominator:
$\dfrac{5-2\sqrt3}{5+2\sqrt3} \times \dfrac{5-2\sqrt3}{5-2\sqrt3}$
$= \dfrac{5^2 - 2(5)(2\sqrt3) + (2\sqrt3)^2}{5^2 - (2\sqrt3)^2}$
$= \dfrac{25 - 20\sqrt3 + 12}{25 - 12}$
$= \dfrac{37 - 20\sqrt3}{13}$
SOLUTION 2
Rewrite the terms inside square roots to match the form $\dfrac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}}$:
$\sqrt{a} = 5 \Rightarrow a = 25$
$\sqrt{b} = 2\sqrt{3} = \sqrt{12} \Rightarrow b = 12$
Apply the shortcut formula $\dfrac{a + b - 2\sqrt{ab}}{a - b}$:
$= \dfrac{25 + 12 - 2\sqrt{25 \times 12}}{25 - 12}$
$= \dfrac{37 - 2\sqrt{300}}{13}$
$= \dfrac{37 - 2(10\sqrt{3})}{13}$
$= \dfrac{37 - 20\sqrt{3}}{13}$
PRACTICE QUESTIONS
Rationalize the denominator.
1. Rationalize the denominator of $\dfrac{7+2\sqrt{3}}{7-2\sqrt{3}}$.
ANSWER: $\dfrac{61+28\sqrt{3}}{37}$
SOLUTION 1
Multiply the numerator and denominator by the conjugate of the denominator:
$\dfrac{7+2\sqrt{3}}{7-2\sqrt{3}}\times\dfrac{7+2\sqrt{3}}{7+2\sqrt{3}}$
$=\dfrac{(7+2\sqrt{3})^2}{7^2-(2\sqrt{3})^2}$
$=\dfrac{49+28\sqrt{3}+12}{49-12}$
$=\dfrac{61+28\sqrt{3}}{37}$
SOLUTION 2
Rewrite the terms to match $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$:
$a=49$
$b=12$
Apply the shortcut formula $\dfrac{a+b+2\sqrt{ab}}{a-b}$:
$=\dfrac{49+12+2\sqrt{49\times12}}{49-12}$
$=\dfrac{61+2(14\sqrt{3})}{37}$
$=\dfrac{61+28\sqrt{3}}{37}$
2. Rationalize the denominator of $\dfrac{5+\sqrt{2}}{5-\sqrt{2}}$.
ANSWER: $\dfrac{27+10\sqrt{2}}{23}$
SOLUTION 1
Multiply the numerator and denominator by the conjugate of the denominator:
$\dfrac{5+\sqrt{2}}{5-\sqrt{2}}\times\dfrac{5+\sqrt{2}}{5+\sqrt{2}}$
$=\dfrac{(5+\sqrt{2})^2}{5^2-(\sqrt{2})^2}$
$=\dfrac{25+10\sqrt{2}+2}{25-2}$
$=\dfrac{27+10\sqrt{2}}{23}$
SOLUTION 2
Rewrite the terms to match $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$:
$a=25$
$b=2$
Apply the shortcut formula $\dfrac{a+b+2\sqrt{ab}}{a-b}$:
$=\dfrac{25+2+2\sqrt{25\times2}}{25-2}$
$=\dfrac{27+2(5\sqrt{2})}{23}$
$=\dfrac{27+10\sqrt{2}}{23}$
3. Rationalize the denominator of $\dfrac{8+\sqrt{3}}{8-\sqrt{3}}$.
ANSWER: $\dfrac{67+16\sqrt{3}}{61}$
SOLUTION 1
Multiply the numerator and denominator by the conjugate of the denominator:
$\dfrac{8+\sqrt{3}}{8-\sqrt{3}}\times\dfrac{8+\sqrt{3}}{8+\sqrt{3}}$
$=\dfrac{(8+\sqrt{3})^2}{8^2-(\sqrt{3})^2}$
$=\dfrac{64+16\sqrt{3}+3}{64-3}$
$=\dfrac{67+16\sqrt{3}}{61}$
SOLUTION 2
Rewrite the terms to match $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$:
$a=64$
$b=3$
Apply the shortcut formula $\dfrac{a+b+2\sqrt{ab}}{a-b}$:
$=\dfrac{64+3+2\sqrt{64\times3}}{64-3}$
$=\dfrac{67+2(8\sqrt{3})}{61}$
$=\dfrac{67+16\sqrt{3}}{61}$
4. Rationalize the denominator of $\dfrac{9-2\sqrt{5}}{9+2\sqrt{5}}$.
ANSWER: $\dfrac{101-36\sqrt{5}}{61}$
SOLUTION 1
Multiply the numerator and denominator by the conjugate of the denominator:
$\dfrac{9-2\sqrt{5}}{9+2\sqrt{5}}\times\dfrac{9-2\sqrt{5}}{9-2\sqrt{5}}$
$=\dfrac{(9-2\sqrt{5})^2}{9^2-(2\sqrt{5})^2}$
$=\dfrac{81-36\sqrt{5}+20}{81-20}$
$=\dfrac{101-36\sqrt{5}}{61}$
SOLUTION 2
Rewrite the terms to match $\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$:
$a=81$
$b=20$
Apply the shortcut formula $\dfrac{a+b-2\sqrt{ab}}{a-b}$:
$=\dfrac{81+20-2\sqrt{81\times20}}{81-20}$
$=\dfrac{101-2(18\sqrt{5})}{61}$
$=\dfrac{101-36\sqrt{5}}{61}$
5. Rationalize the denominator of $\dfrac{10+\sqrt{3}}{10-\sqrt{3}}$.
ANSWER: $\dfrac{103+20\sqrt{3}}{97}$
SOLUTION 1
Multiply the numerator and denominator by the conjugate of the denominator:
$\dfrac{10+\sqrt{3}}{10-\sqrt{3}}\times\dfrac{10+\sqrt{3}}{10+\sqrt{3}}$
$=\dfrac{(10+\sqrt{3})^2}{10^2-(\sqrt{3})^2}$
$=\dfrac{100+20\sqrt{3}+3}{100-3}$
$=\dfrac{103+20\sqrt{3}}{97}$
SOLUTION 2
Rewrite the terms to match $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$:
$a=100$
$b=3$
Apply the shortcut formula $\dfrac{a+b+2\sqrt{ab}}{a-b}$:
$=\dfrac{100+3+2\sqrt{100\times3}}{100-3}$
$=\dfrac{103+2(10\sqrt{3})}{97}$
$=\dfrac{103+20\sqrt{3}}{97}$
6. Rationalize the denominator of $\dfrac{11-2\sqrt{3}}{11+2\sqrt{3}}$.
ANSWER: $\dfrac{133-44\sqrt{3}}{109}$
SOLUTION 1
Multiply the numerator and denominator by the conjugate of the denominator:
$\dfrac{11-2\sqrt{3}}{11+2\sqrt{3}}\times\dfrac{11-2\sqrt{3}}{11-2\sqrt{3}}$
$=\dfrac{(11-2\sqrt{3})^2}{11^2-(2\sqrt{3})^2}$
$=\dfrac{121-44\sqrt{3}+12}{121-12}$
$=\dfrac{133-44\sqrt{3}}{109}$
SOLUTION 2
Rewrite the terms to match $\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$:
$a=121$
$b=12$
Apply the shortcut formula $\dfrac{a+b-2\sqrt{ab}}{a-b}$:
$=\dfrac{121+12-2\sqrt{121\times12}}{121-12}$
$=\dfrac{133-2(22\sqrt{3})}{109}$
$=\dfrac{133-44\sqrt{3}}{109}$