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

`"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`

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

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`