Question

what is the constant term that completes the trinomial

x^2-18x

also I want to know how you got it

Answer 1

You can complete the square for this, I believe

Answer 2

To do that you have to divide the middle term (-18) by 2 and then square it.

-18/2 = 9
9^2 = 81

Answer 3

any number will make a "trinomial"
perhaps you are attempting to complete a square?

Answer 4

are you looking for a perfect square?

in which case you'd want the constant to be 81

essentially when you have a quadratic in the form:

a*x^2 + b*x + c

and you know b (we'll just keep "a" as 1 in this case to keep it simple)

c has to equal (b/2)^2

Thats because when you factor the quadratic to a form of (x+d)^2 where d is a constant,

it expands into x^2 + 2d*x + d^2

now we can again compare this to our original and equate them to solve for our unknowns

original = new
x^2 + b*x + c = x^2 +2*d*x + d^2

the x^2 terms cancel, and you are left with b = 2d and c = d^2

You know b is -18, so now you can solve for d, and then c.

d = b/2 = -18/2 = -9
c=(-9)^2 = 81

so 81 is your constant term

Answer 5

Thanks!