Let $a$ and $b$ with $a>b>0$ be real numbers satisfying $a^2+b^2=4ab$. Find $\dfrac{b}{a}$.
+128408 a^2 + b^2 = 4ab
a^2 / ab + b^2 / ab = 4
a/b + b/a = 4 let a/b = x and b/a =1/x
x + 1/x = 4 multiply through by x
x^2 + 1 = 4x
x^2 - 4x + 1 = 0 complete the square on x
x^2 -4x + 4 = -1 + 4 factor the left, simplify the right
(x - 2)^2 = 3 take both roots
x - 2 = sqrt (3) x - 2 = -sqrt 3
x= sqrt (3) + 2 x = 2 -sqrt (3)
The second solution is < 1.....but since a > b > 0 then a/b must be > 1
So x = sqrt (3) + 2 = a/b
