Hat is the sum of the geometric series in which a1 = 3, r = 4, and an = 49,152? hint: an = a1(r)n − 1, where a1 is the first term and r is the common ratio. sn = −65,535 sn = 16,383 sn = 13,120 sn = 65,535

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)