If p is a negative number and \(0 < s < |p|\), which of the following must also be a negative number?


A) \((p+s)^2\)

B) \((p-s)^2\)

C) \((s-p)^2\)

D) \(p^2-s^2\)

E) \(s^2-p^2\)
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E;

In the number line value of S will be between 0 and +Mod p,

Therefore P2-s2 will always be negative.

Based on the question we know that S is positive, P is negative, but the absolute value of P is greater than S. This means no matter what S is, P squared will be greater than S squared.

The first three answers we can disregard immediately because they are all a sum or difference of S & P but it is squared. So no matter what actual sum or difference is the result squared will be positive.

Now we are left with only D & E as possible solutions to the problem. D is P^2 - S^2, and because the absolute value of P is greater than S, we know that P^2 will be greater than S^2. Example, if S =2, then P is a negative number that is less than -2. So let's assume it's -3.

-3^2 - 2^2 = 9 - 4 = 5. It's positive, and no matter what we substitute in for P & S, as long as the absolute value of P is greater than S, and as such P squared will always be greater than S squared.

The only option left is E.