Question #7b4af

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Question #7b4af

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Answer 1 · 2017-05-23T06:01:39

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

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: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ can be expressed as:
$(x - \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}) (x - \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}) = 0$ $(x-(-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))=0$

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

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

(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:
${(a + b)}^{2}$ $(a+b)^2$ not $(a + b) (c + d)$ $(a+b)(c+d)$

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

So, the discriminant = $0$ $0$ 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 $c=81$

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Question #7b4af

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