If n=-8, what is the smallest perfect square that is greater than n^2 ?

a. 64
b. -64
c. 81
d. -81
e. 16

please answer and explain

81 > n^2 = (-8)^2

a. 64

i use (-8)x (-8) = 64

I use calculator X^2

-8^2 = -64

However, they are asking for the smallest square greater than 64, which is 81.

Note that
-8^2 = -(8^2) = -64
(-8)^2 = +64

To find the smallest perfect square greater than n^2, we first need to find the value of n^2 when n = -8.

n^2 = (-8)^2 = 64

Now, let's determine the smallest perfect square greater than 64.

Starting from the square root of 64, which is 8, we can compare it to the square roots of the given options.

The square root of 64 is already an integer, so we know that 64 is already a perfect square. Therefore, option a. 64 is not the answer.

The square root of -64 is not a real number since we cannot take the square root of a negative number. Therefore, option b. -64 is not the answer.

The square root of 81 is 9, which is greater than the square root of 64. Therefore, option c. 81 is a perfect square greater than 64.

The square root of -81 is not a real number, just like the square root of -64. Therefore, option d. -81 is not the answer.

The square root of 16 is 4, which is smaller than the square root of 64. Therefore, option e. 16 is not the answer.

So, the correct answer is c. 81.