[JEE Advanced 1978] Show that the square of $\dfrac{\sqrt{26-15\sqrt3}}{5\sqrt2-\sqrt{38+5\sqrt3}}$ is rational.
Show that the square of $\dfrac{\sqrt{26-15\sqrt3}}{5\sqrt2-\sqrt{38+5\sqrt3}}$ is rational.
1 Answer
VAVidaara Admin
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· 1mo ago
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Answer: The square equals $\dfrac13$ (rational).
$26-15\sqrt3=\frac{(3\sqrt3-5)^2}2$ and $38+5\sqrt3=\frac{(5\sqrt3+1)^2}2$; the ratio becomes $\frac{3\sqrt3-5}{9-5\sqrt3}$, whose square is $\frac{52-30\sqrt3}{3(52-30\sqrt3)}=\frac13$.
JEE Advanced 1978 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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