How do you find the values of k that will make #9x^2 + 6x +k# a perfect square?

The perfect square is

#9x^2+6x+k=(3x+sqrtk)^2#

#=9x^2+6xsqrtk+k^2#

Comparing the #2# sides

#6sqrtk=6#

#sqrtk=1#

#k=1#

#(3x+a)^2# will give a perfect square

#(3x+a)^2=(3x+a)(3x+a)#

#=9x^2+3ax+3ax+a^2#

#=9x^2+6ax+a^2#

So #9x^2+6ax+a^2=9x^2+6x+k#

#6ax=6x => a=1#

#k=a^2, 1^2=1#

#k=1#

Lets consider a generic case. Say: #(a+b)^2 = a^2+2ab+b^2#

Now lets make a direct comparison between the two:

#(a+b)^2=a^2+2ab+b^2#
# color(white)("dddddddd") 9x^2+6x+k#

So #a^2=9x^2 =>a=sqrt(9x^2) =+-3x#

Also #2ab=6x" so "ab=3x#

As #a=+-3x# then #b=1#

Lets try this out with:#a=+3x#

#(3x+1)(3x+1) = 9x^2+6x+1 color(red)(larr" As required"#

Thus #k=1#

Let , #color(violet)(kinRR # be the #3^(rd) term # to complet square.

#i.e. 9x^2+6x+color(violet)(k)...to(1)#

In the #LHS# we have ,

#color(blue)(diamond 1^(st)term=9x^2#

#color(blue)(diamond2^(nd)term=6x#

#color(blue)(diamond3^(rd)term)=color(violet)(k#

#"We have "color(orange)"formula"# for #3^(rd)term :#

#color(red)(3^(rd)term=(2^(nd)term)^2/(4xx1^(st)term))...to(A)#

#=>color(violet)(k)=(6x)^2/(4xx9x^2)=(36x^2)/(36x^2)#

#=>color(violet)(k=1#

From #(1)#,we get

#9x^2+6x+color(violet)(1)=(3x)^2+2(3x)(1)+(1)^2=(3x+1)^2#

...................................................................................................
Note :

Formula #(A) :color(red)(3^(rd)term=(2^(nd)term)^2/(4xx1^(st)term))# can be use

to find THIRD TERM for any eqn. without any doubt .

WHY ??? #to#Please see below.

#diamond if , a^2+2ab+k=0# ,then [use #(A)#]

#k=(2ab)^2/(4xxa^2)=(4a^2b^2)/(4a^2)=b^2#

#=>a^2+2ab+b^2=(a+b)^2#