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Re: (n+30)^2+(n-30)^2
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29 May 2024, 08:26
Quantity A can be expanded into:
\((n+30)^2+(n-30)^2 = (n^2+60n+900) + (n^2-60n+900) = 2n^2+1800\)
We can see that Quantity A is a quadratic function. At its vertex (n = 0), which is its lowest point, it has a value of 1800. Therefore, Quantity A will always be equal to or greater than 1800.
Quantity B is a linear function, with an intercept of 0. Therefore, Quantity B's value will be negative whenever n is a negative number.
However, we need to check whether there's a value of n where the two functions are equal.
\(2n^2+1800 = 120n\)
\(2n^2-120n+1800 = 0\)
\(2(n^2-60n+900) = 0\)
\(2(n-30)^2 = 0\)
We find that when n = 30, the two are equal to each other. Thus, the relationship cannot be determined.