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

Question

Savannah Anderson · Accepted Answer

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

Genesis Allen · Answer

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

Step 1: Incorporate and eliminate the perfect square element.

The value next to the variable #p#, in this case #3#, is the beginning of the perfect square term. You first cut it in half:

#3 => 3/2#

Next, square the outcome.

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

The original expression should then have this term added and subtracted.

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

Compute the perfect square.

#color(blue)("blue")# contains the perfect square terms.

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

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

Step 3: Reduce the complexity of #color(red)("red")#'s remaining terms.

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

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

Step 4: Determine #p#.

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

Compute #45//4# for each side.

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

Calculate the square root of #+-# for each side.

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

Then deduct #3//2# from each side.

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

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

Abigail Allen · Answer

undefined To solve the equation $ p^2 + 3p - 9 = 0 $ by completing the square, first move the constant term to the other side of the equation: $ p^2 + 3p = 9 $. Then, add the square of half the coefficient of $ p $ to both sides: $ p^2 + 3p + \left(\frac{3}{2}\right)^2 = 9 + \left(\frac{3}{2}\right)^2 $. This simplifies to $ p^2 + 3p + \frac{9}{4} = \frac{45}{4} $. Factor the left side as a perfect square trinomial: $ \left(p + \frac{3}{2}\right)^2 = \frac{45}{4} $. Take the square root of both sides: $ p + \frac{3}{2} = \pm \sqrt{\frac{45}{4}} $. Simplify: $ p + \frac{3}{2} = \pm \frac{3\sqrt{5}}{2} $. Finally, solve for $ p $: $ p = -\frac{3}{2} \pm \frac{3\sqrt{5}}{2} $.