x^2 + 2x - 5 = 0
-
Step 1, make sure the term in front of the x² = 1, if not divide everything by the coefficient to make it equal 1, in your case you already have it equal to 1
x² + 2x - 5 = 0
Step 2, take the coefficient in front of the "x" term (call it b) and add and subtract (b/2)² to the equation. This does not change the equation, because you are adding and subtracting the same term. In your example b=2, so add and subtract (2/2)² = 1:
x² + 2x + 1 - 1 - 5 = 0
Group the "+" term with the x values, and the "-" term with the constant.
(x² + 2x + 1) - 1 - 5 = 0
The items in parenthesis is now a perfect square and factors as (x+(b/2))², combine the constant terms.
(x+1)² - 6 = 0
Solve as necessary:
(x+1)² = 6
x+1 = ±√6
x = -1±√6
x² + 2x - 5 = 0
Step 2, take the coefficient in front of the "x" term (call it b) and add and subtract (b/2)² to the equation. This does not change the equation, because you are adding and subtracting the same term. In your example b=2, so add and subtract (2/2)² = 1:
x² + 2x + 1 - 1 - 5 = 0
Group the "+" term with the x values, and the "-" term with the constant.
(x² + 2x + 1) - 1 - 5 = 0
The items in parenthesis is now a perfect square and factors as (x+(b/2))², combine the constant terms.
(x+1)² - 6 = 0
Solve as necessary:
(x+1)² = 6
x+1 = ±√6
x = -1±√6