How do you find the c that makes the trinomial #x^2+22x+c# a perfect square?

1 Answer
May 15, 2017

Answer:

#c = 121#

Which gives #x^2 +22x+121 =(x+11)^2#

Explanation:

This is a process called 'Completing the Square' and does exactly what the name implies...

To complete means to add what is missing

You are trying to create a perfect square, in this case the square of a binomial.

In #1x^2 + color(red)(b)x + c," "# if this is a perfect square there is always a specific relationship between #b and c#....

'Half of #color(red)(b)#, squared, will give the value of #c#'

This is #c= (color(red)(b)/2)^2#

In #1x^2+ color(red)(22)x + ???" "rarr ??? = (color(red)(22)/2)^2 = 11^2 =121#

The trinomial will therefore be #x^2 +22x+121#, which factorises as

#(x+11)^2#

Note that to do this, the coefficient of #x^2# must be #1#