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

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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]#