Three consecutive even integers are such that the square of the third is 76 more than the square of the second. How do you determine the three integers?

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Three consecutive even integers are such that the square of the third is 76 more than the square of the second. How do you determine the three integers?

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Answer 1 · 2016-01-27T16:19:40

16, 18, and 20.

Explanation:

One can express the three consecuitve even numbers as $2 x, 2 x + 2, and 2 x + 4$ $2x, 2x+2, and 2x+4$ . You are given that ${(2 x + 4)}^{2} = {(2 x + 2)}^{2} + 76$ $(2x+4)^2 = (2x+2)^2 +76$ . Expanding the squared terms yields $4 x^{2} + 16 x + 16 = 4 x^{2} + 8 x + 4 + 76$ $4x^2+16x+16 = 4x^2+8x+4+76$ .
Subtracting $4 x^{2} + 8 x + 16$ $4x^2+8x+16$ from both sides of the equation yields $8 x = 64$ $8x=64$ . So, $x = 8$ $x=8$ . Substituting 8 for x in $2 x, 2 x + 2, and 2 x + 4$ $2x, 2x+2, and 2x+4$ , gives 16,18, and 20.

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Three consecutive even integers are such that the square of the third is 76 more than the square of the second. How do you determine the three integers?

1 Answer

Explanation:

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