Square-root equation

Question

Solve square-root equation: $\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10$

$\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10\\ \left (\sqrt{\left (\sqrt{3}+\sqrt{2} \right )^2} \right )^x+\left (\sqrt{\left (\sqrt{3}-\sqrt{2} \right )^2} \right )^x=10\\ \left (\sqrt{3}+\sqrt{2} \right )^x+\left (\sqrt{3}-\sqrt{2} \right )^x=10$
at the moment I don't know what to do

Note that $\sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3} + \sqrt{2}}$. — Daniel Fischer, Nov 03 '19 at 17:10
$x=2$ is an obvious (eye-balled) solution to the shaded equation, but apparently not the only one — J. W. Tanner, Nov 03 '19 at 17:21

score 6 · Accepted Answer · answered Nov 03 '19 at 17:14

6

Expanding on @DanielFischer's comment, let $y=(\sqrt{3}+\sqrt{2})^x$ so $y+1/y=10$ and $y=5\pm2\sqrt{6}$, so $x=\pm2$.

answered Nov 03 '19 at 17:14

J.G.

115,835

score 1 · Answer 2 · answered Nov 03 '19 at 17:16

1

As an alternative,

quadratic with roots $\alpha=\sqrt{3}+ 2$ and $\beta=\sqrt{3}-2$ is $x^2-2\sqrt{3}x+1=0$

Sum of roots is $\alpha+\beta=2\sqrt{3}$
$\alpha^2+\beta^2=(\alpha+\beta)^2 - 2\alpha\beta = 12-2(1) = 10$

answered Nov 03 '19 at 17:16

AgentS

12,195

Square-root equation

2 Answers2