from the GS:
x/1 = y/x
y = x^2
from the AS:
y-x= 3-y
2y = x+3
y = (x+3)/2
x^2 = (x+3)/2
2x^2 - x - 3 = 0
2x - 3)(x + 1) = 0
x = 2/3 or x = -1
if x = 2/3, then y = 4/9 and
x+y = 2/3 +4/9 = 10/9
if x = -1, then y = 1 and
x+y = 0
so the maximium value of x+y = 10/9
if 1,x,y is a geometric sequence and x,y,3 is an arithmetic sequence, compute the maximum value of x+y
3 answers
y/x = x, so
y = x^2
3-y = y-x, so
2y = x+3
2x^2 = x+3
2x^2 - x - 3 = 0
(x+1)(2x-3) = 0
x = -1 or 3/2
y = 1 or 9/4
x+y = 0 or 15/4
y = x^2
3-y = y-x, so
2y = x+3
2x^2 = x+3
2x^2 - x - 3 = 0
(x+1)(2x-3) = 0
x = -1 or 3/2
y = 1 or 9/4
x+y = 0 or 15/4
Go with Steve's
I was standing on my head when I got
2x-3 = 0 ----> x = 2/3
I was standing on my head when I got
2x-3 = 0 ----> x = 2/3