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.
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)
To determine if the trinomial X^2 + 22X + 121 - y^2 is a perfect square trinomial, we should look for a pattern.
In a perfect square trinomial of the form (a + b)^2, the middle term is twice the product of the square root of the first term and the square root of the last term.
In this case, the first term is X^2, and its square root is X. The last term is 121 - y^2, and its square root is √(121 - y^2).
Using the formula for the middle term, we have: 2 * X * √(121 - y^2).
Since the middle term in the given trinomial is 22X, we can equate it to 2 * X * √(121 - y^2) and solve for X:
22X = 2X * √(121 - y^2).
Dividing both sides of the equation by 2X gives us:
11 = √(121 - y^2).
Now, we can square both sides to solve for (121 - y^2):
(11)^2 = ( √(121 - y^2) )^2
121 = 121 - y^2.
Finally, adding y^2 to both sides gives us:
y^2 = 0.
Therefore, the answer is y = 0.
In conclusion, the trinomial X^2 + 22X + 121 - y^2 is indeed a perfect square trinomial when y = 0.