The sum of squares of two consecutive odd positive integers is 394. Find them.
Answers
SOLUTION :
Let the first consecutive odd positive integer be (x+1) and second consecutive odd positive integer be (x+3).
A .T.Q
(x + 1)² + (x + 3)² = 394
x² + 1 + 2x + x² + 9 + 6x = 394
2x² + 8x + 10 = 394
2x² + 8x + 10 - 394 = 0
2x² + 8x - 384 = 0
2(x² + 4x - 192) = 0
x² + 4x - 192 = 0
x² + 16x - 12x - 192 = 0
x(x + 16) - 12(x + 16) = 0
(x + 16) (x - 12) = 0
(x + 16) = 0 or (x - 12) = 0
x = - 16 or x = 12
Since, x is a odd positive number so x ≠ - 16.
Therefore, x = 12
First odd positive number (x +1) = 12 + 1 = 13
Second odd positive number (x +3) = 12 + 3 = 15.
Hence, the two odd positive numbers be 13 & 15.
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Answer: 13 ; 15
Step-by-step explanation:
Let the first consecutive odd positive integer be (x+1) and second consecutive odd positive integer be (x+3).
A .T.Q
(x + 1)² + (x + 3)² = 394
x² + 1 + 2x + x² + 9 + 6x = 394
2x² + 8x + 10 = 394
2x² + 8x + 10 - 394 = 0
2x² + 8x - 384 = 0
2(x² + 4x - 192) = 0
x² + 4x - 192 = 0
x² + 16x - 12x - 192 = 0
x(x + 16) - 12(x + 16) = 0
(x + 16) (x - 12) = 0
(x + 16) = 0 or (x - 12) = 0
x = - 16 or x = 12
Since, x is a odd positive number so x ≠ - 16.
Therefore, x = 12
First odd positive number (x +1) = 12 + 1 = 13
Second odd positive number (x +3) = 12 + 3 = 15.
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