How do you complete the square for #x^2+18x#?

Question

Eva Adamson · Accepted Answer

#(x+9)^2 = x^2 + 18x +81# Generally speaking, #ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))# Notice that the term added to #x# is #b/(2a)#

Isaiah Anthony · Answer

For a general form, squared binomial
#color(white)("XXXXX")##(x+a)^2 = x^2+2ax+a^2# So if #x^2+18x# are the first two terms of a squared binomial
#color(white)("XXXXX")#then, in the general form, #a=9# and
#color(white)("XXXXX")##a^2 = 9^2 = 81# Of course, if we are going to add #9^2# to the expression #x^2+18x# we are also going to have to subtract it:
#color(white)("XXXXX")##x^2+18x#
#color(white)("XXXXX")##= x^2+18xcolor(red)(+9^2) - color(blue)(9^2)#
#color(white)("XXXXX")##=color(red)((x+9)^2) color(blue)(- 81)#

Eli Assouline · Answer

undefined To complete the square for $ x^2 + 18x $, follow these steps:

1. Factor out the coefficient of $ x^2 $, which is 1.
2. Halve the coefficient of $ x $, which is $ \frac{18}{2} = 9 $.
3. Square the result from step 2, which is $ 9^2 = 81 $.
4. Add and subtract the result from step 3 within the expression: $ x^2 + 18x + 81 - 81 $.
5. Rewrite the expression, grouping the first three terms and factoring the perfect square trinomial: $ (x + 9)^2 - 81 $.

So, the completed square form for $ x^2 + 18x $ is $ (x + 9)^2 - 81 $.