View Full Version : Algebra 1
kidjerk
Jul 30, 2009, 12:35 AM
it says find the value of c that would make the expression a perfect square trinomial the question is x squared minus 6x +c?
ebaines
Jul 30, 2009, 08:46 AM
Well, first of all this is a binomial, not a trinomial. But moving on - you have:
x^2 - 6x + C
and are looking to find C so that this is equivalent to a square:
(x-a)^2
where a is some constant. So multiply this out:
(x-a)^2 = (x-a)*(x-a) = x^2 - 2ax+a^2.
Now look at the middle term - compare it to the original formula and you can determine the value of a, and from that you can find the value of C.
A useful trick to learn about quadratic equations that would serve as a short cut in this problm: if you want a perfect square you take the constant in front of the x term, divide it by 2, and square it - that will give you the value of the constant:
x^2 + bx + b^2/4 = (x+b/2)^2
Edjay
Jul 30, 2009, 08:49 AM
Take 1/2 of the center term and square it. The answer is c=9, and the perfect root is (x-3) squared, that is (x-3)(x-3).
kapilagayan
Jul 30, 2009, 09:06 AM
Lets use this equlity
(x-a)^2=x^2-6x+c
then x^2 -2ax +a^2 = x^2-6x+c
when we consider the coefitiant of each term
6=2a then a=3 then a^2 = 9
since a^2 = c then c=9