find two consecutive positive integers sum of whose square is 365
Answers
Answered by
25
Let the numbers be x and (x+1), then;
x + (x+1) = 365
or, 2x = 364
or, x = 182
Hence, the numbers are 182 and 183.
HOPE THIS COULD HELP!!!
Answered by
15
Answer:
Required numbers are 13 and 14
Step-by-step explanation:
Let two consecutive positive integer be x and x+1
According to your question:
x^2+(x+1)^2 = 365
>>>>x^2+x^2+2x+1 =365
>>>>>2x^2+2x-364 = 0
>>>>>x^2+x - 182 = 0
Let us spilt the middle term,
>>>>x^2+14x-13x-182 =0
>>>>x(x+14) - 13 (x+14) = 0
>>>>> (x+14)(x - 13) = 0
x = - 14 or x = 13
[ - ve root being rejected ]
Hence, required numbers are 13 and 14
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