square root of 2y-1 = 2 + square root y-4

Question

I know that you need to times each side by the 2nd power to get rid of the radical. I am getting

2y-1=4 + y-4   then 0=4 + y -4-2y+1    I got y=1

but the back of the book answer is (5,13)

What step did I not do?

Answer 1

sqrt(2y-1) = 2 + sqrt(y-4)

( sqrt(2y-1))^2 = (2+sqrt(y-4))^2

2y-1 = 4 + 4sqrt(y-4) + y - 4

y-1 = 4sqrt(y-4)

(y-1)^2 = (4sqrt(y-4))^2

y^2 - 2y + 1 = 16(y-4)

y^2 - 2y + 1 = 16y - 64

y^2 - 18y + 65 = 0

A = 1, B = - 18 and C = 65

C = 65 = -3*( - 15)

-3 - 15 = - 18 = B,  and

y^2 - 18y + 65 =0 = (y-3)(y- 15)

y-3 = 0 and y - 15 = 0

y = 3 and y = 15