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

Mar 29, 2018

$+ 9$

#### Explanation:

$\text{to "color(blue)"complete the square}$

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

$\Rightarrow {x}^{2} + 6 x {\textcolor{red}{+ 3}}^{2} = {x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

Mar 29, 2018

${x}^{2} + 6 x + 9 - 9 = {\left(x + 3\right)}^{2} - 9$

#### Explanation:

To complete square one is basically doing

${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$
or
${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$

We can see that ${x}^{2} = {a}^{2}$ and
$2 a b = 6 x$

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

So if we put the ${b}^{2}$ term in we get

${x}^{2} + 6 x + 9 - 9 = {\left(x + 3\right)}^{2} - 9$

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

Mar 29, 2018

${x}^{2} + 6 x + \textcolor{red}{9} = {\left(x + 3\right)}^{2}$

#### Explanation:

We have,

x^2+6x+square?.

First Term $= F . T . = {x}^{2}$

MiddleTerm $= M . T . = 6 x$

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,

${x}^{2} + 6 x + \textcolor{red}{9} = {\left(x + 3\right)}^{2}$

e.g.

$\left(1\right) {a}^{2} + 2 a b + \textcolor{red}{{b}^{2}} = {\left(a + b\right)}^{2}$

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

$\left(2\right) a + 2 \sqrt{a b} + \textcolor{red}{b} = {\left(\sqrt{a} + \sqrt{b}\right)}^{2}$

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