1) wat term should be added to x squared+22x, so tat the result is a perfect square trrinomial
I think that is 1/2 the b term (1/2*22 and that squared) so the term to be added would be 11^2 = 121.
x^2 + 22x + 121 = (x+11)(x+11)
(x + a)(x + a)
=x^2 + 2ax + a^2
In the problem,
2ax=22x
2a=22
a=11
substituting back
(x + 11)(x + 11)
x^2 + 22x + 121
To determine the term that should be added to make the expression x² + 22x a perfect square trinomial, we need to recognize that perfect square trinomials have the form (x + a)², where "a" is some constant.
In order to find "a," we need to take half of the coefficient of x in the given expression (in this case, 22) and square it (multiply it by itself).
Step 1: Take half of the coefficient of x: 22/2 = 11
Step 2: Square the result: 11² = 121
Therefore, the term that needs to be added to x² + 22x to make it a perfect square trinomial is 121.
The resulting expression will be: x² + 22x + 121, which can be factored as (x + 11)².