How do you solve #p^2 + 3p - 9 = 0# by completing the square?

2 Answers
May 28, 2017

#p=-3/2-sqrt(45)/2# and #p=-3/2+sqrt(45)/2#

Explanation:

Step 1. Add and subtract the perfect square term.

The perfect square term starts with the value next to the variable #p#, in this case #3#. First, you cut it in half:

#3 => 3/2#

Then you square the result

#3/2 => (3/2)^2=9/4#

Then add and subtract this term in the original expression.

#p^2+3p+9/4-9/4-9=0#

Step 2. Factor the perfect square.

The terms in #color(blue)("blue")# are the perfect square.

#color(blue)(p^2+3p+9/4)-9/4-9=0#

#color(blue)((p+3/2)^2)-9/4-9=0#

Step 3. Simplify the remaining terms in #color(red)("red")#.

#(p+3/2)^2color(red)(-9/4-9)=0#

#(p+3/2)^2color(red)(-45/4)=0#

Step 4. Solve for #p#.

#(p+3/2)^2-45/4=0#

Add #45//4# to both sides

#(p+3/2)^2=45/4#

Take #+-# the square root of both sides.

#p+3/2=+-sqrt(45/4)#

Subtract #3//2# from both sides.

#p=-3/2+-sqrt(45)/2#

#p=-3/2-sqrt(45)/2# and #p=-3/2+sqrt(45)/2#

May 28, 2017

#p=-3/2+-[3sqrt(5)]/2#

Explanation:

#p^2+ 3p-9=0# => add 9 to both sides:
#p^2+3p=9# => using;#(a+b)^2=a^2+ 2ab+b^2#
#p^2+3p+(3/2)^2= 9 + 9/4#
#(p+3/2)^2=45/4# => take square root of both sides:
#p+3/2=+-sqrt(45)/2# => simplify:
#p=-3/2+-[3sqrt(5)]/2#