Final answer:
To find all pairs of positive integers that satisfy x^4 + y^2 < 100, we determine the maximum possible integer values for x, then find the corresponding y values for each x. The total number of satisfying pairs is 20.
Step-by-step explanation:
The question asks how many pairs of positive integers satisfy the inequality x^4 + y^2 < 100. To solve this, we must identify the integer values of x and y that, when put in the inequality, will result in a sum less than 100.
Let's consider the possible values of x first, since the fourth power grows very quickly. x can be 1, 2, or 3 because 1^4=1, 2^4=16, and 3^4=81 are all less than 100, while 4^4=256 is too large.
For each value of x, we now find the possible values of y:
- If x=1, y can be any value from 1 to 9 because 1+81=82 is the maximum value for y^2.
- If x=2, y can range from 1 to 9 as well, because 16+81=97 is less than 100.
- If x=3, y can only be 1 or 2, since 3^4 + 2^2 = 81 + 4 = 85, and 3^4 + 3^2 = 81 + 9 = 90, both are less than 100.
Thus, the total number of pairs is 9 (for x=1) + 9 (for x=2) + 2 (for x=3) = 20 pairs.