Algebra: Solve x for [ {2 + (3^1/2)}^1/2]^x + [{2 – (3^1/2)}^1/2]^x = 4?

The trick is this. SQUARE THE EQUATION.
Note that : [{2 + (3^1/2)}^1/2]^x = [{2 + (3^1/2)}^(x/2)]

We get

[ {2 + (3^1/2)}^1/2]^x + [{2 – (3^1/2)}^1/2]^x = 4

{2 + (3^1/2)}^x
+ 2 [{2 + (3^1/2)}^(x/2)]*[{2 – (3^1/2)}^(x/2)]
+ {2 – (3^1/2)}^x = 16

and [{2 + (3^1/2)}^(x/2)]*[{2 – (3^1/2)}^(x/2)] equals to
[ 4 – 3 ] ^(x/2) = 1.

Thus, we get,
{2 + (3^1/2)}^x + 2 + {2 – (3^1/2)}^x = 16
{2 + (3^1/2)}^x + {2 – (3^1/2)}^x = 14 …(1)

Now note that
[{2 + (3^1/2)}^(x)]*[{2 – (3^1/2)}^(x)] = [4 – 3]^x = 1
means that,
{2 – (3^1/2)}^x = 1 / [{2 + (3^1/2)}^(x)]
Substituiting this into (1), we get,
{2 + (3^1/2)}^x + 1 / [{2 + (3^1/2)}^(x)] = 14

Let y = [{2 + (3^1/2)}^(x)]
The equation above becomes,

y + 1 / y = 14
y^2 – 14y + 1 = 0

Which gives y = [14 ± sqrt(192)]/2

Since y = [{2 + (3^1/2)}^(x)], Thus,

[{2 + (3^1/2)}^(x)] = [14 ± sqrt(192)]/2
= [14 ± 8sqrt(3)]/2
= 7 ± 4*3(1/2)
So,
x = ln [ 7 ± 4*3^(1/2)] / ln [2 + (3^1/2)]
= ± 2

Alternately,
One can use the identity,
{2 + (3^1/2)}^(2) = 7 + 4*3(1/2)

{2 + (3^1/2)}^(-2) = 7 – 4*3(1/2)

To get x = ± 2 again.

Done.

The below is ADDITIONAL information solving a slightly different problem though similar to the one you posted.

Say we have a similar problem but x is a positive real.
We want to find solutions for
[ 2 + {(3^1/2)}^1/2]^x + [2 – {(3^1/2)}^1/2]^x = 4

By observation, x = 1 is a solution since,

[ {2 + (3^1/2)}^1/2] + [{2 – (3^1/2)}^1/2] = 4

How to show they arent any other solutions?

( x > 0 )

Let f(x) =[ {2 + (3^1/2)}^1/2]^x + [{2 – (3^1/2)}^1/2]^x – 4
So, our equation amounts for solutions for f(x) = 0.
We show that f(x) is increasing for x>0 and we are done.
Note that when we show f(x) increasing, it means that
f(x)=0 only at one value because for any x > y, increasing means f(x)> f(y).
Since f(1) = 0, Thus,
f(y) > f(1) = 0 for y > 1 and
f(y) < f(1) = 0 for 0 < y < 1

All that remains to be shown is that f(x) is increasing.
Simplest way, is to use basic calculus.
Its necessary and sufficent to show f'(x)>0 for x >0.
Using the fact for constants any constant m, if y = m^x,
Then dy/dx = (m^x)(ln m)

f'(x)= [{2 + (3^1/2)}^1/2]^x *ln[{2 + (3^1/2)}^1/2]
+ [{2 – (3^1/2)}^1/2]^x *ln[{2 – (3^1/2)}^1/2]

Now, note that
[{2 + (3^1/2)}^1/2] = 3.316074013
[{2 – (3^1/2)}^1/2] = 0.683925987
ln[{2 + (3^1/2)}^1/2] = 1.198781557
ln[{2 – (3^1/2)}^1/2] = – 0.379905573

So,
f'(x) = (3.316074013^x)*(1.198781557)
– (0.683925987^x)*(0.379905573)

So to show f'(x) >0, its equivalent to showing,

(3.316074013^x)*(1.198781557) >(0.683925987^x)*(0.379905573)
(4.848586069)^x > (0.316909757)
x ln (4.848586069) > ln (0.316909757)
1.57868713 x > -1.149138224
x > – 0.727907513

which is true for all x > 0.
Thus, for x>0, the only solution to our modified
problem is x=1.