Final answer:
The three consecutive whole numbers whose squares sum up to 434 are 7, 8, and 9. We found these numbers by setting up an equation for the squares of the numbers, expanding it, and then factorizing the quadratic equation.
Step-by-step explanation:
We are looking for three consecutive whole numbers whose squares sum up to 434. Let us denote the smallest of these numbers as n. Therefore, the numbers we are looking for can be represented as n, n+1, and n+2. The problem then becomes finding n such that n^2 + (n+1)^2 + (n+2)^2 = 434.
Expanding the equation gives us n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 = 434. Simplifying this, we get 3n^2 + 6n + 5 = 434. Subtracting 434 on both sides gives 3n^2 + 6n - 429 = 0. Factorizing the quadratic equation, we find that (3n - 21)(n + 21) = 0. This gives us two possible values for n, but only n = 7 is appropriate since n cannot be negative. Thus, the three consecutive whole numbers are 7, 8, and 9.