Sum of squares of 2 successive positive integers is 481. Sum of integers would be

Bawa   14 years AGO

Sum of squares of 2 successive positive integers is 481. If x is one integer, then ....
x^2 +(x+1)^2 = 481 ; 2x^2 +2x=480
x(x+1) =240; then x= 15.......
Sum of integers would be 15+16=31

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Dr.Dipin Singh   14 years AGO

Sum of squares of 2 successive positive integers is 481. Average of squares of 2 successive positive integers is 481/2= 240.5.

.....

Hence value of square of one integer is less than 240.5 and value of square of one integer is more than 240.5 . Then possible values are 15 and 16 such that

15^2+16^2=481 and 15+16=31..option 4

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KALPANA G   14 years AGO

two positive successive numbers are x and x+1.
according to the problem, x^2|+(x+1)^2=481
==>x^2+x-480=0
==>x=15 or -16
x is positive. so, x=15.then two numbers are 15 and 16.
sum of the integers=15+16=31

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Khanna   14 years AGO

x^2+x^2+2x+1=481 ;
2x^2+2x-480=0;
x^2+x-240=0;
x(x+1) =240; 15*16=240, then 15+16=31

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Saurabh Dua   14 years AGO

let one number be x.
so x^2+(x+1)^2=481
we get x=15.
so 15+16=31

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Raushan Kumar   14 years AGO

15^2+16^2=225+256=481

sum of integer=15+16=31

option=4

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Subodh Kumar   14 years AGO

15*15+16*16=481

so,,,, 15+16=31

option=4

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SUDESHNA SEN   14 years AGO

x^2+(x-1)^2= 481
or, x^2+x^2-2x+1= 481
or, 2x^2-2x-480=0
or, x^2-x-240=0
or, x^2-16x+15x-240=0
or, x(x-16)+15(x-16)=0
or, (x-16)(x+15)=0
So, x= 16, -15
Since x is positive, so x=16.
So, the integers are 16 and 15 and their sum is 16+15= 31..... Option 4)

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Achelal Sharma   14 years AGO

the numbers got by forming the equation x2 + (x+1)2= 481
is 16 & 15

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Mihir   14 years AGO

Sum of squares of 2 successive positive integers ( 15 and 16) is 481. Sum of integers 15 and 16 would be 31.

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Nishat Parveen   14 years AGO

4. 31

480/2=240
now we have sqrt(240)=15.sth
hence, we have 15 and 16 as the nos.
15+16=31

cross check:
15^2+16^2=481

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ravi jha   14 years AGO

15^2+16^16=481
+
so answer is 15+16=31

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Karthik Pandiyan   14 years AGO

Let x be first integer and successive integer be x+1
Given as : x^2+(x+1)^2=481
2x^2+2x-480=0
x=15 0r -16
Since x must be +ve so x=15 and x+1=16
Sum of 2 integers =31

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suneil   14 years AGO

x^2 + y^2 = 481
x + y = ?
x = 15 & y = 16
15^2 + 16^2 = 481
15 + 16 = 31 (option 4)

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goutam Choudhury   14 years AGO

x^2+(x+1)^=481,or 2x^2+2x+1=481,or2x^2+2x-480=0, or x^2+x-240=0, or (x-15)(x+16)=0. So x=15.Negative value ignored. The result is 15^2+16^2=481. Number is 15+16=31.

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HARMEET KAUR   14 years AGO

15+16=31

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Rama Chandra Prakash Kapa   14 years AGO

(15*15)+(16*16) =481
So the successive positive integers are 15 and 16.
So the sun of successive positive integers are 15+16=31

hence option 4 is correct

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Suresh Varma P   14 years AGO

15^2 + 16^2 = 481
so, 15 + 16 = 31

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dharmaveer   14 years AGO

x^2 + y^2 = 481 gives us x^2 + x = 250. i.e., x=16.
it is provided that "Sum of squares of 2 successive positive integers is 481".
and the first number is 15. so, the next number = 15+1=16

so Sum of integers is 15 + 16 =31

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Mamon Sengupta   14 years AGO

x^2+(x-1)^2= 481
or, x^2+x^2-2x+1= 481
or, 2x^2-2x-480=0
or, x^2-x-240=0
or, x^2-16x+15x-240=0
or, x(x-16)+15(x-16)=0
or, (x-16)(x+15)=0
So, x= 16, -15
Since x is positive, so x=16.
So, the integers are 16 and 15 and their sum is 16+15= 31..... Option 4)

Like? Yes (1) | No (1)  

Trishita Dasgupta   14 years AGO

x^2+(x-1)^2= 481
or, x^2+x^2-2x+1= 481
or, 2x^2-2x-480=0
or, x^2-x-240=0
or, x^2-16x+15x-240=0
or, x(x-16)+15(x-16)=0
or, (x-16)(x+15)=0
So, x= 16, -15
Since x is positive, so x=16.
So, the integers are 16 and 15 and their sum is 16+15= 31..... Option 4)

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Sahil   14 years AGO

15+16=31

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zoheb hashmi   14 years AGO

15^2=225 & 16^=256 225+256=481

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Abhishek Divakar   14 years AGO

let integers are a and a+1.
then a^2 + (a+1)^2 = 481
a^2 + a^2 + 1 + 2a = 481
2 (a^2 + a) = 480
a^2 + a - 240 = 0
(a + 16) (a - 15) = 0
a = 15 0r a = -16
a = 15 (as a is Positive)
integers are 15 and 16.
sum is 31.

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Biswajyoti Biswas   14 years AGO

two successive numbers: x,x+1
given,x^2+(x+1)^2=481
or,(x-15)(x+16)=0
as,x is a +ve integer then x=15 obviously
so,sum of successive integers= x+(x+1)=15+16=31

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Ranjana Sinha   14 years AGO

say the 2 integers are a and (a+1).
then,a^2+(a+1)^2=481
=>2a^2+2a-480=0

hence we get,a=-16 ,a=15.
given in the question, the integers are positive.
so, a=15
so the two integers are 15 and 16.
so the sum is 15+16=31

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Ravi B Mali   14 years AGO

x^2 + (x+1)^2 = 481
x^2 + x - 480 = 0
solving we get x = 15

so sum of positive integers x and x+1 is 15+16= 31

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