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11 votes
11 votes
Find three consecutive whole numbers such that the sum of the squares of the numbers is equal to 434

User StuiterSlurf
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3.1k points

2 Answers

27 votes
27 votes

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.

User Globalfish
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2.7k points
15 votes
15 votes

Answer:


\left \{ {{x^2+y^2+z^2=434} \atop {x-y-z=0}} \right.

Step-by-step explanation:

User Tobias Schula
by
3.0k points