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Find two consecutive positive integers, sum of whose squares is 365. [ncert 10th.]
Let the two consecutive positive integers
are x and x+1.
Given that x²+(x+1)² = 365
Using (a+b)=a²+b²+2ab, we get,
=> x²+x²+1+2x=365.
=> 2x²+2x+1-365=0
=> 2x²+2x-364=0
Dividing both sides by 2 (non zero), we get,
=> x²+x-182=0
=> x²+14x-13x-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
=> Either x= -14 , 13
But we have to find positive integer, so -14 must be rejected.
So x= 13 is the required positive integer.
Now two consecutive positive integers shall be x, x+1
=> 13 and 13+1=14
So 13 and. 14 are the required positive consecutive integers. Ans.
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