If the arithmetic mean of x and y is equal to the product of sqrt(2) and the geometric mean of x and y, then find y/x.
+128474 [x + y] / 2 = √2 * √(xy) square both sides
[ x^2 + 2xy + y^2 ] / 4 = 2 (xy)
x^2 + 2xy + y^2 = 8 (xy)
x^2 + y^2 = 6(xy) divide through by x^2
1 + (y^2) / (x^2) = 6(y/x) rearrange as
(y^2/x^2) - 6(y/x) + 1 = 0
(y/x)^2 - 6(y/x) + 1 = 0
Let (y/x) = a
a^2 - 6a + 1 = 0
a^2 - 6a = -1 complete the square on a
a^2 -6a + 9 = -1 + 9
(a - 3)^2 = 8 take both roots
a -3 = ±√ 8
a - 3 = ±2√2
a = 3 ± 2√2
(y/x) = 3 ± 2√2
