How do you complete the square to solve #x^2 + 5x + 6 = 0#?

1 Answer
Jun 16, 2015

You can complete the square by first getting the form #x^2 + kx = h#.

#x^2 + 5x = -6#

Then just add and subtract a certain value that is equal to #(k/2)^2#. Just remember that the function is #x^2 + kx#, so #k# may be negative, but the added #(k/2)^2# will always be positive.

#x^2 + 5x + (5/2)^2 = (5/2)^2 - 6#

#(x+5/2)^2 = 25/4 - 24/4 = 1/4#

#(x+5/2)^2 - 1/4 = 0#

graph{(x+5/2)^2 - 1/4 [-10, 10, -5, 5]}

If you wanted to solve this:

#(x+5/2)^2 = 1/4#

#x+5/2 = pmsqrt(1/4)#

Thus:
#x+5/2 = pmsqrt(1/4)#

#x = pm(sqrt(1/4)) - 5/2#

#x = pm(1/2) - 5/2#

#x = 1/2 - 5/2 = -2#

#x = -1/2 - 5/2 = -3#

#(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0#
#(-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0#

You could just factor, though...

#(x+2)(x+3) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6#