the square of the binomial x+1 on hundred and twenty greater than the square of the binomial x-3?
We can start by using the formula for the square of a binomial:
(a + b)^2 = a^2 + 2ab + b^2
In this case, we have:
(x + 1)^2 = x^2 + 2x + 1
and
(x - 3)^2 = x^2 - 6x + 9
So the problem can be written as:
x^2 + 2x + 1 + 120 = x^2 - 6x + 9
Simplifying this equation, we get:
8x = -112
Dividing both sides by 8, we get:
x = -14
Therefore, the solution is x = -14.
(x+1)^ - (x-3)^2 = 120
x^2 + 2x + 1 - x^2 + 6x - 9 = 120
8x = 128
x = 16
check:
if x = 16, (x+1)^2 = 289
(x-3)^2 = 169
since 289 > 169 by 120 , my answer is right, and the bot is wrong