If $a$ and $b$ are integers, such that $a\not= 0$ and $b\not= 0$ and $a^2$ and $b^2$ have at most two digits, what is the greatest possible difference between the squares of $a$ and $b?$

80

Step-by-step explanation:

t is important to note that a square of any non-zero integer is positive, and therefore there is no advantage in using negative integers instead of positive integers to attain the greatest difference of squares. So we will not consider negative integers.

The greatest value of a^2 - b^2 occurs when a^2 is at its largest and b^2 is at its smallest.

The larger a, the larger a^2:

8 ^ 2 = 64

9 ^ 2 = 81

10 ^ 2= 100

Since a^2 can have at most two digits, a=10 is too large, and so a=9 is the largest integral value of a we can use.

Now, b^2 is at its smallest when b is closest to zero on the number line (the further b gets from zero, the larger its square becomes):

2 ^ 2 = 4

1 ^ 2 =1

0 ^ 2 = 0

Remember to go back to the original problem sometimes, to make sure you are taking everything into account. It states b doesn't =0, and therefore the b=1 is the closest b can get to zero as an integer. So, the greatest difference between b^2 and a^2 is when b=1 and a=9, giving the result:

a^2-b^2 =9^2-1^2 =81-1= 80.

So, 80 is your answer.

the answer is c hope this