﻿ 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}} }

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 3 months ago

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 3 months ago

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