Question #7b4af

1 Answer
May 23, 2017

Answer:

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#