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`