# Question #7b4af

May 23, 2017

The completed square: ${x}^{2} + 18 x + 81$
Factorized: ${\left(x - 9\right)}^{2}$

#### Explanation:

In order for this to be a perfect square, there must only be one, and only one root.
The reason for this is that any quadratic: $a {x}^{2} + b x + c = 0$ can be expressed as:
$\left(x - \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(x - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right) = 0$

In this case, the two roots of x would be:
$\frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}$ and $\frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}$

Since:
$x - \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} = 0$ and $x - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a} = 0$

(You get the point)

Notice how there are 2 terms in the factorized equation. In order to complete a perfect square, we must follow this format:
${\left(a + b\right)}^{2}$ not $\left(a + b\right) \left(c + d\right)$

In order for that to happen, $\frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}$

So, the discriminant = $0$ or $\Delta = 0$ or ${b}^{2} - 4 a c = 0$

Now, we can plug this in and solve, using ${x}^{2} + 18 x$, $a = 1$ and $b = 18$:

${b}^{2} - 4 a c = 0$
$\Rightarrow {18}^{2} - 4 \cdot 1 \cdot c = 0$
$\Rightarrow 4 c = {18}^{2}$
$\Rightarrow c = 81$

Since $\Delta = 0$, then $x = - \frac{b}{2 a}$, $b = 18 , a = 1$
So, $x = 9$

Then we can piece this all together:
${x}^{2} + 18 x + 81 = {\left(x + 9\right)}^{2}$

We have completed this square with $c = 81$