Multiple choice

The number of real roots of the equation $(\sqrt{3}+1)^{2x}+(\sqrt{3}-1)^{2x}=2^{3x}$ is equal to

  1. $0$
  2. $1$
  3. $2$
  4. more than $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Divide by (sqrt(3)+1)^(2x). The equation becomes 1 + ((sqrt(3)-1)/(sqrt(3)+1))^(2x) = (2/(sqrt(3)+1))^(2x). This simplifies to 1 + (2-sqrt(3))^x = (sqrt(3)-1)^(2x). Testing x=1 gives 1 + 2-sqrt(3) = 3-2sqrt(3)+1, which is not equal. Testing x=0 gives 1+1=1 (false). Graphical analysis shows only one intersection.

AI explanation

Rewrite the equation as (4+2sqrt(3))^x + (4-2sqrt(3))^x = 8^x, then divide by 8^x to get ((4+2sqrt(3))/8)^x + ((4-2sqrt(3))/8)^x = 1. The fractional bases simplify to (1+sqrt(3))/4 and (1-sqrt(3))/4, which can be rewritten as (1/2)cos(30) ± (1/2)sin(30). Thus the equation is (1/2)^x * (cos(30) + sin(30))^x + (1/2)^x * (cos(30) - sin(30))^x = 1, which simplifies to (1/2)^x * 2cos(30x) = 1 or 2cos(30x) = 2^x. By inspection, x = 0 is one real root (2 = 1), and for x > 0 the increasing function 2^x quickly surpasses the oscillating value of 2cos(30x), yielding exactly one real root.

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