Multiple Choice Question (MCQ)

What is the value of the radical expression

\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6} }-\dfrac{1}{\sqrt{5-2\sqrt{6}} }

  1. ×

    2

  2. ×

    -1

  3. ×

    \sqrt{2}+\sqrt{3}

  4. Non of the above

Collected in the board: Simplify Radical Expressions

Steven Zheng posted 1 hour ago

Answer

  1. Observed \sqrt{2}, \sqrt{3} and \sqrt{6} appears in both terms of the expression.

    Expand the square of \sqrt{2}+\sqrt{3} for a study.

    (\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}

    Then the first term could be simplified as

    \sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6} }

    =\sqrt{(\sqrt{2}+\sqrt{3} )^2+2(\sqrt{2}+\sqrt{3})+1 }

    =\sqrt{(\sqrt{2}+\sqrt{3} +1)^2}

    =\sqrt{2}+\sqrt{3} +1

    Multiply numerator and denominator with the conjugate of the radicant. The second term is simplified by using of difference of squares formula.

    \dfrac{1}{\sqrt{5-2\sqrt{6}} }

    =\dfrac{1}{\sqrt{5-2\sqrt{6} }}\cdot \dfrac{\sqrt{5+2\sqrt{6}}}{\sqrt{5+2\sqrt{6}}}

    =\sqrt{5+2\sqrt{6}}

    =(\sqrt{\sqrt{2}+\sqrt{3} )^2}

    =\sqrt{2}+\sqrt{3}

    Then, subtracting the second term from the first term gives

    \sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6} }-\dfrac{1}{5-2\sqrt{6} }

    =\sqrt{2}+\sqrt{3} +1 - (\sqrt{2}+\sqrt{3} )

    =1

    Therefore, none of the choices are correct. D is the right choice.

Steven Zheng posted 1 hour ago

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