How do you find the value of c that makes #x^2-22x+c# into a perfect square?

1 Answer
Nov 23, 2016

Answer:

#c=121 #

Explanation:

We have

#x^2 -22x + c#

We want to make this into a perfect square, and suppose we choose the value of k such that:

# x^2 -22x + c = (x+k)^2 # (NB we expect k to -ve)
# :. x^2 -22x + c = x^2+2k+k^2 #

Comparing coefficient of #x# we have

#2k=-22 => k=-11 #

And comparing constant coefficients we have:
# c=k^2 => c=121 #

Hence If we choose #c=121# then we can write

# x^2 -22x + 121 = (x-11)^2 #