Show that the square of \frac{\left(\right. \sqrt{26 - 15 \sqrt{3}} \left.\right)}{\left(\right. 5 \sqrt{2} - \sqrt{38 + 5 \sqrt{3}} \left.\right)} is a rational number.

Foundation

Mathematics Foundation

Question

Show that the square of

\frac{(\sqrt{26 - 15 \sqrt{3}})}{(5 \sqrt{2} - \sqrt{38 + 5 \sqrt{3}})}

is a rational number.

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Given,

x = \frac{\sqrt{26 - 15 \sqrt{3}}}{5 \sqrt{2} - \sqrt{38 + 5 \sqrt{3}}}

x^{2} = \frac{26 - 15 \sqrt{3}}{50 + 38 + 5 \sqrt{3} - 10 \sqrt{76 + 10 \sqrt{3}}}

\Rightarrow x^{2} = \frac{26 - 15 \sqrt{3}}{88 + 5 \sqrt{3} - 10 \sqrt{75 + 1 + 10 \sqrt{3}}}

\Rightarrow x^{2} = \frac{26 - 15 \sqrt{3}}{88 + 5 \sqrt{3} - 10 \sqrt{(5 \sqrt{3} + 1)^{2}}}

\Rightarrow x^{2} = \frac{26 - 15 \sqrt{3}}{88 + 5 \sqrt{3} - 10 (5 \sqrt{3} + 1)}

\Rightarrow x^{2} = \frac{26 - 15 \sqrt{3}}{3 (26 - 15 \sqrt{3})} = \frac{1}{3}

which is a rational number.