How do you solve $y^2 - 3y - 14 = 0$ using completing the square?

Question

2023 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-06T12:03:52

Add 14 to both sides, then add half the square of the coefficient of $y$ to both sides, then solve.

1) Add 14 to both sides of the equation

$y^2-3y=14$

2) Add half the square of $y$ 's coefficient to both sides

$y^2-3y + (-3/2)^2 = 14 + (-3/2)^2$

3) Now the left side is a perfect square; remember to simplify the right hand side too

$(y-3/2)^2=65/4$

4) Take $+-$ the square root of both sides; remember the left side is a square, so it remains positive (i.e., it's not $+-$ )

$y-3/2=+-sqrt(65/4)$

5) Add the remaining fraction on the left to both sides and simplify your roots

$y=3/2+-sqrt(65)/2$

The solution set is: $[3/2-sqrt(65)/2, 3/2+sqrt(65)/2]$

How do you solve y^2 - 3y - 14 = 0 y^2 - 3y - 14 = 0 using completing the square?