Question
Find the two consecutive positive integers, such that the sum of their squares is 365.
Answer: Option B
:
B
Let the numbers be x&x+1
Square of consecutive number: x2&(x+1)2
Sum of square:x2+(x+1)2
∴x2+(x+1)2=365
Lets solve
x2+x2+2x+1=3652x2+2x−364=0x2+x−182=0
(Dividing by 2)
Factorise,
x2+14x−13x−182=0x(x+14)−13(x+14)=0(x+14)(x−13)=0∴x=13&−14
Since numbers are positive integers, x=13
So, required consecutive numbers are 13&14.
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:
B
Let the numbers be x&x+1
Square of consecutive number: x2&(x+1)2
Sum of square:x2+(x+1)2
∴x2+(x+1)2=365
Lets solve
x2+x2+2x+1=3652x2+2x−364=0x2+x−182=0
(Dividing by 2)
Factorise,
x2+14x−13x−182=0x(x+14)−13(x+14)=0(x+14)(x−13)=0∴x=13&−14
Since numbers are positive integers, x=13
So, required consecutive numbers are 13&14.
Was this answer helpful ?
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