How would one complete the square: #x^2 + 6x +# _?

3 Answers
Mar 29, 2018

Answer:

#+9#

Explanation:

#"to "color(blue)"complete the square"#

#• " add "(1/2"coefficient of the x-term")^2" to"#
#x^2+6x#

#rArrx^2+6xcolor(red)(+3)^2=x^2+6x+9=(x+3)^2#

Mar 29, 2018

Answer:

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

Explanation:

To complete square one is basically doing

#a^2+2ab+b^2=(a+b)^2#
or
#a^2-2ab+b^2=(a-b)^2#

We can see that #x^2=a^2# and
#2ab=6x#

So all we need to condense this into #(a+b)^2# is a #b^2# term
We know that
#2b=6# as #x=a#
so #b=3#
and #b^2=9#

So if we put the #b^2# term in we get

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

We include the #+-9# because we have to net add nothing to the equation so #9-9=0# so we really haven't added anything

Mar 29, 2018

Answer:

#x^2+6x+color(red)(9)=(x+3)^2#

Explanation:

We have,

#x^2+6x+square?.#

First Term # =F.T.=x^2#

MiddleTerm #=M.T.=6x#

Third Term# =T.T.=square?#

Let us use the Formula:

#color(red)(T.T.=(M.T.)^2/(4xx(F.T.))=(6x)^2/(4xx(x^2))=(36x^2)/(4x^2)=9#

Hence,

#x^2+6x+color(red)(9)=(x+3)^2#

I think no need to double check the answer.Please see below.

e.g.

#(1)a^2+2ab+color(red)(b^2)=(a+b)^2#

#T.T.=(2ab)^2/(4xxa^2)=(4a^2b^2)/(4a^2)=color(red)(b^2#

#(2)a+2sqrt(ab)+color(red)(b)=(sqrta+sqrtb)^2#

#T.T.=(2sqrt(ab))^2/(4xxa)=(4ab)/(4a)=color(red)(b#

#(3)613089x^2+1490832xy+color(red)(906304y^2)= (783x+952y)^2#

#T.T.=(1490832xy)^2/(4xx613089x^2)= (2222580052224x^2y^2)/(2452356x^2)=color(red)(906304y^2#