find 2 consecutive integer sum of those sq is 365
Answers
Answer:
hopefully it will help you
Step-by-step explanation:
The next consecutive positive integer will be x + 1.
According to the given question,
x² + ( x + 1)² = 365
x² +( x + 1)² = 365
x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²]
2x² + 2x + 1 = 365
2x² + 2x + 1- 365 = 0
2x² + 2x - 364 = 0
2(x² + x - 182) = 0
x² + x - 182 = 0
x² + 14x - 13x - 182 = 0
x (x + 14) - 13 (x + 14) = 0
(x - 13) (x + 14) = 0
x - 13 = 0 and x + 14 = 0
x = 13 and x = - 14
The value of x cannot be negative (because it is given that the integers are positive).
∴ x = 13 and x + 1 = 14
Answer:
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Step-by-step explanation:
Let the consecutive positive integers be x and x + 1.
Therefore, x2 + (x + 1)2 = 365
⇒ x2 + x2 + 1 + 2x = 365
⇒ 2x2 + 2x - 364 = 0
⇒ x2 + x - 182 = 0
⇒ x2 + 14x - 13x - 182 = 0
⇒ x(x + 14) -13(x + 14) = 0
(x + 14)(x - 13) = 0
Either x + 14 = 0 or x - 13 = 0,
⇒ x = - 14 or x = 13
Since the integers are positive, x can only be 13.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.