Math, asked by mjsabiq, 20 days ago

the sum of the squares of two consecutive odd integers is 290 then find these integers​

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Answered by shambhukumar141006
0

Answer:

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Answered by Acer27
1

Answer: ± 11 and ± 13

Step-by-step explanation:

Let the consecutive odd digits be (2n + 1) and (2n + 3),

according to the question :-

(2n + 1)^{2} + (2n + 3)^{2} = 290

We know the identity :- (a + b)² = a² + b² + 2ab

We therefore get the following equation :-

→ 4n² + 1 + 4n + 4n² + 9 + 12n = 290

→ 8n² + 16n = 280

→ 8n ( n + 2) = 280

→ n + 2 = 280 / 8n

→ n + 2 = 35/n

⇒ n² + 2n - 35 = 0

We can now factorize the following equation to find out its factors :-

n² + 7n - 5n - 35 = 0

→  n (n + 7) - 5 (n + 7) = 0

⇒ (n + 7) (n - 5) = 0

We get the value of n to be either 5 or -7.

Now putting these values into the equation :-

(2n + 1)^{2} + (2n + 3)^{2} = 290

i) When n = 5

→ (10 + 1)² + (10 + 3)² = 121 + 169 = 290

The values of the integers could be 11 and 13.

ii) When n = -7

→ (-14 + 1)² + (-14 + 3)² = 169 + 121 = 290

The values of the integers could b -11 and -13 as well.

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