Question

Question #7b4af

Answer 1

The completed square: `x^2+18x+81`
Factorized: `(x-9)^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: `ax^2+bx+c=0` can be expressed as:
`(x-(-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))=0`

In this case, the two roots of x would be:
`(-b+sqrt(b^2-4ac))/(2a)` and `(-b-sqrt(b^2-4ac))/(2a)`

Since:
`x-(-b+sqrt(b^2-4ac))/(2a)=0` and `x-(-b-sqrt(b^2-4ac))/(2a)=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:
`(a+b)^2` not `(a+b)(c+d)`

In order for that to happen, `(-b-sqrt(b^2-4ac))/(2a)=(-b+sqrt(b^2-4ac))/(2a)`

So, the discriminant = `0` or `Delta=0` or `b^2-4ac=0`

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

`b^2-4ac=0`
`rArr18^2-4*1*c=0`
`rArr4c=18^2`
`rArrc=81`

Since `Delta=0`, then `x=-b/(2a)`, `b=18, a=1`
So, `x=9`

Then we can piece this all together:
`x^2+18x+81=(x+9)^2`

We have completed this square with `c=81`