For example, if you are calculating 144÷36{\displaystyle {\sqrt {144}}\div {\sqrt {36}}}, rewrite the problem like this: 14436{\displaystyle {\frac {\sqrt {144}}{\sqrt {36}}}}.
For example, 14436{\displaystyle {\frac {\sqrt {144}}{\sqrt {36}}}} can be rewritten as 14436{\displaystyle {\sqrt {\frac {144}{36}}}}.
For example, 14436=4{\displaystyle {\frac {144}{36}}=4}, so 14436=4{\displaystyle {\sqrt {\frac {144}{36}}}={\sqrt {4}}}.
For example, 4 is a perfect square, since 2×2=4{\displaystyle 2\times 2=4}. Thus:4{\displaystyle {\sqrt {4}}}=2×2{\displaystyle ={\sqrt {2\times 2}}}=2{\displaystyle =2}So, 14436=4=2{\displaystyle {\frac {\sqrt {144}}{\sqrt {36}}}={\sqrt {4}}=2}.
For example, if you are calculating 8÷36{\displaystyle {\sqrt {8}}\div {\sqrt {36}}}, rewrite the problem like this: 836{\displaystyle {\frac {\sqrt {8}}{\sqrt {36}}}}.
For example:836=2×2×26×6{\displaystyle {\frac {\sqrt {8}}{\sqrt {36}}}={\frac {\sqrt {2\times 2\times 2}}{\sqrt {6\times 6}}}}
For example: 2×2×26×6{\displaystyle {\frac {\sqrt {{\cancel {2\times 2\times }}2}}{\sqrt {\cancel {6\times 6}}}}} 226{\displaystyle {\frac {2{\sqrt {2}}}{6}}}So, 836=226{\displaystyle {\frac {\sqrt {8}}{\sqrt {36}}}={\frac {2{\sqrt {2}}}{6}}}
For example, if your expression is 623{\displaystyle {\frac {6{\sqrt {2}}}{\sqrt {3}}}}, you need to multiply the numerator and denominator by 3{\displaystyle {\sqrt {3}}} to cancel the square root in the denominator:623×33{\displaystyle {\frac {6{\sqrt {2}}}{\sqrt {3}}}\times {\frac {\sqrt {3}}{\sqrt {3}}}}=62×33×3{\displaystyle ={\frac {6{\sqrt {2}}\times {\sqrt {3}}}{{\sqrt {3}}\times {\sqrt {3}}}}}=669{\displaystyle ={\frac {6{\sqrt {6}}}{\sqrt {9}}}}=663{\displaystyle ={\frac {6{\sqrt {6}}}{3}}}.
For example, 26{\displaystyle {\frac {2}{6}}} reduces to 13{\displaystyle {\frac {1}{3}}}, so 226{\displaystyle {\frac {2{\sqrt {2}}}{6}}} reduces to 123{\displaystyle {\frac {1{\sqrt {2}}}{3}}}, or simply 23{\displaystyle {\frac {\sqrt {2}}{3}}}.
For example, if you are calculating 432616{\displaystyle {\frac {4{\sqrt {32}}}{6{\sqrt {16}}}}}, you would first simplify 46{\displaystyle {\frac {4}{6}}}. The numerator and denominator can both be divided by a factor of 2. So, you can reduce: 46=23{\displaystyle {\frac {4}{6}}={\frac {2}{3}}}.
For example, since 32 is evenly divisible by 16, you can divide the square roots:3216=2{\displaystyle {\sqrt {\frac {32}{16}}}={\sqrt {2}}}.
For example, 23×2=223{\displaystyle {\frac {2}{3}}\times {\sqrt {2}}={\frac {2{\sqrt {2}}}{3}}}.
For example, if your expression is 4327{\displaystyle {\frac {4{\sqrt {3}}}{2{\sqrt {7}}}}}, you need to multiply the numerator and denominator by 7{\displaystyle {\sqrt {7}}} to cancel the square root in the denominator:437×77{\displaystyle {\frac {4{\sqrt {3}}}{\sqrt {7}}}\times {\frac {\sqrt {7}}{\sqrt {7}}}}=43×77×7{\displaystyle ={\frac {4{\sqrt {3}}\times {\sqrt {7}}}{{\sqrt {7}}\times {\sqrt {7}}}}}=42149{\displaystyle ={\frac {4{\sqrt {21}}}{\sqrt {49}}}}=4217{\displaystyle ={\frac {4{\sqrt {21}}}{7}}}
For example, if you are calculating 15+2{\displaystyle {\frac {1}{5+{\sqrt {2}}}}}, you have a binomial in the denominator, since 5+2{\displaystyle 5+{\sqrt {2}}} is a two-termed polynomial.
For example, 5+2{\displaystyle 5+{\sqrt {2}}} and 5−2{\displaystyle 5-{\sqrt {2}}} are conjugate pairs, since they have the same terms but opposite operations.
For example:15+2{\displaystyle {\frac {1}{5+{\sqrt {2}}}}}=1(5−2)(5+2)(5−2){\displaystyle ={\frac {1(5-{\sqrt {2}})}{(5+{\sqrt {2}})(5-{\sqrt {2}})}}}=5−2(52−(2)2{\displaystyle ={\frac {5-{\sqrt {2}}}{(5^{2}-({\sqrt {2}})^{2}}}}=5−225−2{\displaystyle ={\frac {5-{\sqrt {2}}}{25-2}}}=5−223{\displaystyle ={\frac {5-{\sqrt {2}}}{23}}}Thus, 15+2=5−223{\displaystyle {\frac {1}{5+{\sqrt {2}}}}={\frac {5-{\sqrt {2}}}{23}}}.
