Question

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

Answer 1

`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`