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Question

Find the consecutive positive integers, sum of whose square is 365.

Solution
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There is difference of 1 in consecutive positive integers.

Let first integer =x
Second integer =x+1
Also given that
Sum of squares =365

x2+(x+1)2=365
x2+x2+1+2x=365
2x2+1+2x=365
2x2+1+2x365=0
2x2+2x364=0
2(x2+x182)=0
x2+x182=0
x2+14x13x182=0
x(x+14)13(x+14)=0
(x13)(x+14)=0
x13=0 x+14=0

x=13 x=14

So, the root of the equation are x=13 & x=14

Since we have to find consecutive positive number

We take x=13

First number =x=13

Second number =x+1=13+1=14.

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