How do you complete the square for #3x^2 +18x + 5#?

Question

7260 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-05T08:49:03

#3(x+3)^2-22#

#3x^2+18x+5#

take out common factor to make coefficient #x^2=1#

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

now CTS as normal inside the bracket as follows

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

#=3((x+3)^2-9)+5#

#=3(x+3)^2-27+5#

#=3(x+3)^2-22#

Answer 2 · 2018-04-05T08:55:05

#3(x + 3)^2 - 22#

Before completing the square a must equal 1, so first we divide all terms by 3.

= #(3x^2)/3 + (18x)/3 + (5)/3#

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

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

= #3((x^2 + 6x + 9) - 9 + 5/3)#

= #3((x^2 + 6x + 9) - 27/3 + 5/3)#

= #3((x^2 + 6x + 9) - 22/3)#

#(x^2 + 2xy + y^2) = (x + y)^2#, so...

= #3((x + 3)^2 - 22/3)#

= #3(x + 3)^2 - 22#