How do you find the c that makes the trinomial #x^2+9x+c# a perfect square?

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Answer 1 · 2017-04-30T09:07:10

#c=81/4#

Note that:

#(x+b/2)^2 = x^2+bx+b^2/4#

So if #b=9# then:

#c = b^2/4 = color(blue)(9)^2/4 = 81/4#

#color(white)()#
Bonus

More generally, we find:

#ax^2+bx+c = a(x+b/(2a))^2+(c-b^2/(4a))#

which if #a=1# simplifies to:

#x^2+bx+c = (x+b/2)^2+(c-b^2/4)#

as in our example.

Answer 2 · 2017-04-30T09:13:18

The pattern is:

#(x+a)^2=x^2+2ax+a^2#

Match the terms of the given equation with the right side of the pattern.

Match the terms of the given equation, #x^2+9x+c# with the right side of the pattern.

Matching the first terms:

#x^2=x^2# does not help us.

Matching the second terms gives us the following equation:

#2ax=9x#

We can use the above equation to solve for the value of "a" by dividing both sides of the equation by 2x:

#a=9/2#

Matching the constant terms, gives us the following equation:

#a^2 = c#

Substitute #9/2# for "a":

#(9/2)^2 = c#

This is your answer:

#c = 81/4#