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Q.

Find two consecutive odd positive integers, sum of whose squares is 290.


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a

9 and 7

b

11 and 13

c

5 and 7

d

1 and 3 

answer is B.

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Detailed Solution

Suppose one of the even positive numbers = x.
Other odd positive integer = x+2
Now sum of their square is 290. So,
290=x2+(x+2)2 
290=x2+x2+4x+4 
290=2x2+4x+4  2x2+4x=286 
2x2+4x-286=0
x2+2x-143=0
x2+13x-11x-143=0 
xx+13-11x+13=0
(x-11)=0,(x+13)=0  Thus, x=11 or -13
Neglect negative value..
Here, x=11 and
(x+2)=11+2=13.
So, the odd positive integers are 11 and 13.
Correct option is 2.
 
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