the sum of the squares of two consecutive odd integers is 290 then find these integers
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Answer: ± 11 and ± 13
Step-by-step explanation:
Let the consecutive odd digits be (2n + 1) and (2n + 3),
according to the question :-
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 :-
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.