Square root of \sqrt{13

Foundation

Mathematics Foundation

Surds

Algebraic Identities

Rationalisation

Question

Square root of

\sqrt{13 - 4 \sqrt{3}}

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Let

\sqrt{13 - 4 \sqrt{3}} = \sqrt{x} - \sqrt{y}

\Rightarrow 13 - 4 \sqrt{3}, x + y - 2 \sqrt{x y}

\Rightarrow x + y = 13, x y = 12

\Rightarrow x^{2} + y^{2} = 169 - 24 = 145

\Rightarrow {(x - y)}^{2} = 145 - 24 = 121 \Rightarrow x - y = 11

∴ x = 12, y = 1

∴ \sqrt{13 - 4 \sqrt{3}} = \sqrt{12} = 1

∴ \sqrt{\sqrt{13 - 4 \sqrt{3}}} = \sqrt{\sqrt{12} - 1}

Hence, solved.