Find the sum of the real values of x such that the infinite geometric series \(x+\frac{1}{2}x^3+\frac{1}{4}x^5+\frac{1}{8}x^7+\dots\) is equal to -12.
+36916 a1 = x r = 1/2 x^2
Sn = a1 (1-r^n)/(1-r) = a1/(1-r) = x / ( 1 - 1/2x^2) = -12
x = -12 + 6x^2
0 = 6x^2 -x - 12 x = 18/12 and -16/12
added together = 2/12 = 1/6
