Originally Posted by

**topsquark** **x^2 - 2x - 5 = 0**

x^2 - 2x = 5

Now, we want to add a number to both sides such that the LHS will be a perfect square. As it happens x^2 + 2a + a^2 = (x + a)^2. Comparing the coefficient of the linear term of this form with the linear term in the equation above we can see that

2a = -2

a = -1

Thus a^2 = (-1)^2 = 1. So add 1 to both sides:

x^2 - 2x + 1 = 5 + 1

(x - 1)^2 = 6

x - 1 = (+/-)sqrt(6)

x = 1 (+/-) sqrt(6)

We can check this using the quadratic formula:

x^2 - 2x - 5 = 0

Gives

x = {-(-2) (+/-) sqrt([-2]^2 - 4*1*(-5))}/(2*1)

x = {2 (+/-) sqrt(24)}/2 = {2 (+/-) sqrt(4*6)}/2 = {2 (+/-) 2*sqrt(6)}/2

x = 1 (+/-) sqrt(6) (Check!)

-Dan