Question

find two consecurive positive integers 'the sum of whose squares is 365.

Answer 1

When writing what you have given and taking smaller as x, it looks like x^2+(x+1)^2=365
→x^2+x^2+2x+1=365
→2x^2+2x=364
→x^2+x=182
→x^2+x-182=0
→x^2+14x-13x-182=0
→x(x+14)-13(x+14)=0
→(x+14)(x-13)=0
→x=-14 or x=13
since we need positive integers
So x=13
And x+1=14

So the no. you are looking for are →13,14

Hope it will help you

Answer 2

Hi friend ,

Let the positive integer be x
Second integer = x+1

ATQ
(x+1)² +x² = 365

x² +1² + 2×x×1 + x² = 365

x² +1+2x+x² = 365

2x² +1+2x = 365

2x²+1+2x-365 = 0

2x² -364 +2x = 0

2x²+2x-364 = 0.................(1)

2x²+(28-26)x -364 = 0

2x²+28x-26x-364 = 0

2x(x+14)-26(x+14) = 0

(x+14)(2x-26) = 0

x +14 = 0

x = -14

2x-26 = 0

2x = 26

x = 26/2

x = 13

Since positive integers are never in negative ,-14 is rejected. One integer = 13
Second integer = x+1 = 13+1 = 14

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Hope this helped u...