Asked by Nelson
I need help with this problem, and I think it's a perfect square trinomial problem.
X^2+22+121-y^2
I know 11 is the square of 121.
X^2+22+121-y^2
I know 11 is the square of 121.
Answers
Answered by
Damon
22x maybe?
what is the question anyway?
factor it?
put in an equal sign and solve it
find zeros of parabola ?
by the way
x^2 + 22 x + 121 = (x+11)^2
(x+11)^2 -y^2 = (x+11-y)(x+11+y)
because
a^2-b^2 = (a-b)(a+b)
what is the question anyway?
factor it?
put in an equal sign and solve it
find zeros of parabola ?
by the way
x^2 + 22 x + 121 = (x+11)^2
(x+11)^2 -y^2 = (x+11-y)(x+11+y)
because
a^2-b^2 = (a-b)(a+b)
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