Let S=x R: (√3+√2)^x^2-4+(√3-√2)^x^2-4=10.

Question

Let $S=\left\{x\in\mathbb R: (\sqrt3+\sqrt2)^{x^2-4}+(\sqrt3-\sqrt2)^{x^2-4}=10\right\}$. Then $n(S)$ is equal to

Accepted Answer

Let $t=(\sqrt3+\sqrt2)^{x^2-4}$. Since $(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=1$, we have $(\sqrt3-\sqrt2)^{x^2-4}=1/t$.

$$t+\frac1t=10 \implies t^2-10t+1=0 \implies t=5\pm2\sqrt6.$$

Since $(\sqrt3+\sqrt2)^2=5+2\sqrt6$ and $(\sqrt3+\sqrt2)^{-2}=(\sqrt3-\sqrt2)^2=5-2\sqrt6$:

$t=5+2\sqrt6$: $x^2-4=2 \implies x^2=6 \implies x=\pm\sqrt6$.

$t=5-2\sqrt6$: $x^2-4=-2 \implies x^2=2 \implies x=\pm\sqrt2$.

So $S=\{\pm\sqrt6,\pm\sqrt2\}$, and $n(S)=4$.

Solutions · 1