[JEE Advanced 1978] show that the square of sqrt 26 15 sqrt 3 5 sqrt 2 sqrt
Show that the square of $\dfrac{\sqrt{26-15\sqrt3}}{5\sqrt2-\sqrt{38+5\sqrt3}}$ is rational.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
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.
Log in to post your own answer or join the discussion.
No comments yet — start the discussion.