Solve by completing the square.
y^2 - 2y - 27 = 0
Write your answer as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundreth.
y = ? or y = ?
To complete the square, we need to add and subtract the square of half of the coefficient of the y term.
First, we can factor out a 1 from the y^2 - 2y term:
y^2 - 2y - 27 = 0
y^2 - 2y = 27
Next, take half of the coefficient of the y term (-2/2 = -1) and square it (1):
y^2 - 2y + 1 - 1 = 27
The "+1 -1" doesn't change the value of the equation, it's just a way to add 0 to the equation so we can complete the square.
Now we can rewrite the left side as a perfect square:
(y - 1)^2 - 28 = 0
Add 28 to both sides:
(y - 1)^2 = 28
Take the square root of both sides (remembering to include both positive and negative square roots):
y - 1 = ±√28
y = 1 ± 2√7
So the solutions are:
y = 1 + 2√7
y = 1 - 2√7