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keke0529
18.11.2020 •
Mathematics
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Ответ:
A. 32
Step-by-step explanation:
Ответ:
"The sq. of the 3rd is 7 greater than the sum of the squares of the first two" translates into:
(x+4)^2 = 7 + (x)^2 + (x+2)^2
Expanding, x^2 + 8x + 16 = 7 + x^2 + x^2 + 4x + 4
Cross out x^2 on both sides: 8x + 16 = 7 + x^2 + 4x + 4
Subtract 16 from both sides: 8x = 7 + x^2 + 4x + 4 -16
or: 8x = x^2 + 4x + 11 - 16
8x = x^2 + 4x - 5
Subtracting 8x from both sides: 0 = x^2 -4x - 5
Solve this by completing the square:
0 = x^2 - 4x + 4 -4 - 5, or
0 = (x-2)^2 - 9
Thus, x-2 = plus or minus sqrt(9)
x-2 = plus or minus 3
x = 2 plus or minus 3
x = 5 or x = -1
Try x = 5. Then the consec. odd integers are {5, 7, 9} (answer)