How do you solve by completing the square for #2x^2+7x - 15 = 0#?

1 Answer
Mar 30, 2015

Group the two variable terms together, then factor out the number in front of #x^2#

#(2x^2+7x)-15=0#

#2(x^2+7/2 x+ color(white)"leave space") -15=0#

Take #1/2# of the coefficient of #x#, square that and add and subtract inside the parentheses:

#1/2 " of " 7/2 = 7/4# Squaring gives us #49/16#, so we write:

#2(x^2+7/2 x+ 49/16 - 49/16) -15=0# Keep the positive #49/16# inside the parentheses (we need it for the perfect square)

Write:
#2(x^2+7/2 x+ 49/16) - 2(49/16) -15=0#

Factor the square and simplify the rest:

#2(x + 7/4)^2 - 49/8 - 120/8=0#

#2(x + 7/4)^2 - 169/8=0#

#2(x + 7/4)^2 = 169/8#

#(x + 7/4)^2 = 169/16#

# x + 7/4 = +- sqrt(169/16)#

# x + 7/4 = +- 13/4#

# x = -7/4 +- 13/4#

# -7/4 + 13/4 = 6/4=3/2# and # -7/4 - 13/4 = -20/4 = -5#

The solutions are #3/2#, #5#