Question 170210
find three consecutive odd integers such that the squar of the thid integer plus the product of the other two integers is 268.
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Let x = first consecutive odd integer
then
x+2 = second consecutive odd integer
x+4 = third consecutive odd integer
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From:"the square of the third integer plus the product of the other two integers is 268" we get:
(x+4)^2 + x(x+2) = 268
x^2+8x+16 + x^2+2x = 268
2x^2 + 10x + 16 = 268
x^2 + 5x + 8 = 134
x^2 + 5x - 126 = 0
(x-9)(x+14) = 0
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x = {9, -14}
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We can toss out the -14, leaving us with:
x=9
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Solution:
9, 11, 13