The numbers are 16, 17, and 18.
Given statement,
three consecutive whole numbers such that the sum of the squares of the numbers is equal to 869.
Let's assume that the first number be x, then the next consecutive numbers will be, (x+1), (x+2), (x+3),......
We only need three numbers so,
x² + (x+1)² + (x+2)² = 869
expanding the brackets,
x²+x²+1²+2x+x²+4+4x = 869
3x² + 6x + 5 =869
3x² + 6x + 5 -869 =0
3x² + 6x -864 = 0
Dividing by 3,
x² + 2x - 288 = 0
We know that,when ax² + bx + c = 0, factors of x are,
[tex]x = \dfrac{-b\pm\sqrt{b^2 -4ac}}{2a}[/tex]
putting in the values,
[tex]x = \dfrac{-2\pm\sqrt{2^2 -4\times 1\times -288}}{2\times 1}[/tex]
x = 16, -18
As -18 is a negative value, 16 is the only option left.
x = 16
x+1 = 17
x+2 = 18
Trying 16, 17, and 18.
16²+17²+18² = 869,
Hence, the numbers are 16, 17, and 18.
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