find two
consecutive positive integers of whose square is 365
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hope it helps...i'd tried mah best....
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Answered by
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I guess your question is to find two consecutive positive numbers, the "sum" of squares of which is 365
If it is so,then here is the solution...
Let the two consecutive numbers be x and x+1
Therefore,according to the question:-
x²+(x+1)²=365
x²+x²+1²+2*1*x=365
2x²+1+2x=365
2x²+2x=365-1
2x²+2x=364
2(x²+x)=364
x²+x=364/2=182
x²+x-182=0
x²+14x-13x-182=0
x(x+14)-13(x+14)=0
(x+14) (x-13)=0
If x+14=0
Then x=-14,which is negative hence doesn't fit in the question
Hence x-13=0
That gives,x=13
Now that x=13,the other number is x+1=13+1=14
Hence,the numbers are 13 and 14
Hope this helps!
If it is so,then here is the solution...
Let the two consecutive numbers be x and x+1
Therefore,according to the question:-
x²+(x+1)²=365
x²+x²+1²+2*1*x=365
2x²+1+2x=365
2x²+2x=365-1
2x²+2x=364
2(x²+x)=364
x²+x=364/2=182
x²+x-182=0
x²+14x-13x-182=0
x(x+14)-13(x+14)=0
(x+14) (x-13)=0
If x+14=0
Then x=-14,which is negative hence doesn't fit in the question
Hence x-13=0
That gives,x=13
Now that x=13,the other number is x+1=13+1=14
Hence,the numbers are 13 and 14
Hope this helps!
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