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

1 Answer
May 25, 2016

Answer:

#x=-3/2+-(3sqrt(5))/2#

#x~~1.854" "and" " -4.854 #

Explanation:

Not that standard form is #y=ax^2+bx+c#

and that vertex form is #y=color(red)(a)(x+color(red)(b/(2a)))^2+c +k#

where #k# is a correction constant in your case #a=1#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "y=x^2+3x-9=0#

write as: #0=(x^2+3x)-9#

Add the correction constant #k#

#0=(x^2+3x)-9+k#

Remove the #x# from #3x#

#0=(x^2+3)-9+k#

Halve the 3

#0=(x^2+3/2)-9+k#

Move the power from #x^2# to outside the bracket

#0=(x+3/2)^2-9+k" "color(red)(larr "the error comes from "a(3/(2a))^2)#

but #k+(3/2)^2=0 => k=-9/4# giving

#color(blue)("Vertex form "->0=(x+3/2)^2- 45/4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Take the #-45/4# to the other side of the equals and change its sign.

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

Square root both sides

#x+3/2=sqrt(45/4)#

Move the #3/2# to the other side of =

#x=-3/2+-sqrt(5xx9)/sqrt(4)#

#color(blue)(x=-3/2+-(3sqrt(5))/2)#

#color(blue)(x~~1.854" "and" " -4.854 ) #

Tony B