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Rationalize the denominator of the following:

Mathematics

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6 parts
i

2−32-\sqrt{3}

ii

3−223-2\sqrt{2}

iii

43+523\frac{4\sqrt{3}+5}{23}

iv

323−161132\sqrt{3}-16\sqrt{11}

v

83−18279\frac{83 - 18\sqrt{2}}{79}

vi

12+14312 + \sqrt{143}

Step-by-step solution

19 steps

Here are the solutions for rationalizing the denominator for each expression:

1. Rationalize the denominator of the following:

Part i· 3 steps

12+3\frac{1}{2+\sqrt{3}}

  1. Multiply the numerator and denominator by the conjugate of the denominator, which is 2−32-\sqrt{3}.

    12+3×2−32−3\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}

  2. Apply the difference of squares formula (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2-b^2 in the denominator.

    2−3(2)2−(3)2=2−34−3\frac{2-\sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2-\sqrt{3}}{4-3}

  3. Simplify the expression.

    2−31=2−3\frac{2-\sqrt{3}}{1} = 2-\sqrt{3}

    The rationalized expression is 2−3\boxed{2-\sqrt{3}}.

Part ii

3−223-2\sqrt{2}

13+22\frac{1}{3+2\sqrt{2}}

  1. Multiply the numerator and denominator by the conjugate of the denominator, which is 3−223-2\sqrt{2}.

    13+22×3−223−22\frac{1}{3+2\sqrt{2}} \times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}

  2. Apply the difference of squares formula in the denominator.

    3−22(3)2−(22)2=3−229−(4×2)\frac{3-2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2} = \frac{3-2\sqrt{2}}{9 - (4 \times 2)}

  3. Simplify the expression.

    3−229−8=3−221=3−22\frac{3-2\sqrt{2}}{9-8} = \frac{3-2\sqrt{2}}{1} = 3-2\sqrt{2}

    The rationalized expression is 3−22\boxed{3-2\sqrt{2}}.

Part iii

43+523\frac{4\sqrt{3}+5}{23}

143−5\frac{1}{4\sqrt{3}-5}

  1. Multiply the numerator and denominator by the conjugate of the denominator, which is 43+54\sqrt{3}+5.

    143−5×43+543+5\frac{1}{4\sqrt{3}-5} \times \frac{4\sqrt{3}+5}{4\sqrt{3}+5}

  2. Apply the difference of squares formula in the denominator.

    43+5(43)2−(5)2=43+5(16×3)−25\frac{4\sqrt{3}+5}{(4\sqrt{3})^2 - (5)^2} = \frac{4\sqrt{3}+5}{(16 \times 3) - 25}

  3. Simplify the expression.

    43+548−25=43+523\frac{4\sqrt{3}+5}{48-25} = \frac{4\sqrt{3}+5}{23}

    The rationalized expression is 43+523\boxed{\frac{4\sqrt{3}+5}{23}}.

Part iv

323−161132\sqrt{3}-16\sqrt{11}

1623+11\frac{16}{2\sqrt{3}+\sqrt{11}}

  1. Multiply the numerator and denominator by the conjugate of the denominator, which is 23−112\sqrt{3}-\sqrt{11}.

    1623+11×23−1123−11\frac{16}{2\sqrt{3}+\sqrt{11}} \times \frac{2\sqrt{3}-\sqrt{11}}{2\sqrt{3}-\sqrt{11}}

  2. Apply the difference of squares formula in the denominator.

    16(23−11)(23)2−(11)2=16(23−11)(4×3)−11\frac{16(2\sqrt{3}-\sqrt{11})}{(2\sqrt{3})^2 - (\sqrt{11})^2} = \frac{16(2\sqrt{3}-\sqrt{11})}{(4 \times 3) - 11}

  3. Simplify the expression.

    16(23−11)12−11=16(23−11)1=16(23−11)\frac{16(2\sqrt{3}-\sqrt{11})}{12-11} = \frac{16(2\sqrt{3}-\sqrt{11})}{1} = 16(2\sqrt{3}-\sqrt{11}) =323−1611= 32\sqrt{3}-16\sqrt{11}

    The rationalized expression is 323−1611\boxed{32\sqrt{3}-16\sqrt{11}}.

Part v

83−18279\frac{83 - 18\sqrt{2}}{79}

9−29+2\frac{9-\sqrt{2}}{9+\sqrt{2}}

  1. Multiply the numerator and denominator by the conjugate of the denominator, which is 9−29-\sqrt{2}.

    9−29+2×9−29−2\frac{9-\sqrt{2}}{9+\sqrt{2}} \times \frac{9-\sqrt{2}}{9-\sqrt{2}}

  2. Expand the numerator using (a−b)2=a2−2ab+b2(a-b)^2 = a^2-2ab+b^2 and the denominator using (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2-b^2.

    (9−2)2(9)2−(2)2=92−2(9)(2)+(2)281−2\frac{(9-\sqrt{2})^2}{(9)^2 - (\sqrt{2})^2} = \frac{9^2 - 2(9)(\sqrt{2}) + (\sqrt{2})^2}{81 - 2}

  3. Simplify the expression.

    81−182+279=83−18279\frac{81 - 18\sqrt{2} + 2}{79} = \frac{83 - 18\sqrt{2}}{79}

    The rationalized expression is 83−18279\boxed{\frac{83 - 18\sqrt{2}}{79}}.

Part vi

12+14312 + \sqrt{143}

13+1113−11\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}

  1. Multiply the numerator and denominator by the conjugate of the denominator, which is 13+11\sqrt{13}+\sqrt{11}.

    13+1113−11×13+1113+11\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}} \times \frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}}

  2. Expand the numerator using (a+b)2=a2+2ab+b2(a+b)^2 = a^2+2ab+b^2 and the denominator using (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2-b^2.

    (13+11)2(13)2−(11)2=(13)2+2(13)(11)+(11)213−11\frac{(\sqrt{13}+\sqrt{11})^2}{(\sqrt{13})^2 - (\sqrt{11})^2} = \frac{(\sqrt{13})^2 + 2(\sqrt{13})(\sqrt{11}) + (\sqrt{11})^2}{13 - 11}

  3. Simplify the expression.

    13+2143+112=24+21432\frac{13 + 2\sqrt{143} + 11}{2} = \frac{24 + 2\sqrt{143}}{2}

  4. Factor out 2 from the numerator and simplify.

    2(12+143)2=12+143\frac{2(12 + \sqrt{143})}{2} = 12 + \sqrt{143}

    The rationalized expression is 12+143\boxed{12 + \sqrt{143}}.

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