[JEE Advanced 1985] solve for x 5 2 sqrt 6 x 2 3 5 2 sqrt 6
Solve for $x$: $(5+2\sqrt6)^{x^2-3}+(5-2\sqrt6)^{x^2-3}=10$.
1 Answer
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· 2d ago
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Answer: $x=\pm2,\ \pm\sqrt2$.
The bases are reciprocals; with $u=(5+2\sqrt6)^{x^2-3}$, $u+\frac1u=10\Rightarrow u=5\pm2\sqrt6=(5+2\sqrt6)^{\pm1}$, so $x^2-3=\pm1\Rightarrow x^2=4$ or $2$.
JEE Advanced 1985 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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