4. Find two consecutive positive integers, sum
of whose squares is 365.
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4. Find two consecutive positive integers, sum
of whose squares is 365.
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Answer:
The first number = 13
The other consecutive number = 14
Step-by-step explanation:
Let the first number = x
Let the other consecutive number = x+1
According to the question :-
(x)^2 + (x+1)^2 = 365
=> x^2 + x^2 +2x +1 = 365
=> 2(x^2 + x) = 364
=> x^2 +x = 182
=> x^2 + x - 182 = 0
=> x^2 + 14x - 13x - 182 = 0
=> x(x+14) -13 (x + 14) =0
=> (x-13)(x+14) = 0
Hence, the value of x is -14 and 13
Number can only be positive as mentioned in the question
So,
The first number = x = 13
The other consecutive number = x+1 = 13 + 1 = 14
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ANSWER :-
We Have To calculate the two consecutive positive integers whose squares sum is 365.
So,
Let the first positive integers be 'x'
& other be '(x+1)'
According to question -
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