find twoconsecutive odd positive integers sum of whose squares 202
Answers
Answer:
The two consecutive odd positive integers are: 9 and 11
Explanation:
Let "x" be the first odd integer.
Hence, the consecutive odd integer will be "x + 2"
ATQ,
x² + (x + 2)² = 202
x² + (x² + 4x + 4) = 202
2x² + 4x + 4 = 202
2x² + 4x = 198
x² + 2x = 99
x² + 2x - 99 = 0
(x + 11)(x - 9) = 0
x = -11 or x = 9
As the integer is positive, x = -11 is not admissible.
=> x = 9 and (x + 2) = 11
The two consecutive odd positive integers are 9 and 11
Verify:
Sum of squares of the given numbers = 9² + 11² = 81 + 121 = 202 (verified)
Answer:
9 and 11
Explanation:
Since they are consecutive odd integers, then we can just divide 202 by 2, square root it, then round to the nearest two odd integers.
In this case, 202 / 2 = 101 ⇒ 1010.5 ≈ 10.04.
The two nearest odd integers are 9 and 11. We can verify that this is the pair:
9^2 + 11^2
=81 + 121 = 202.