Please don’t spam answer: find all sets of three consecutive odd integers such that the sum of the squares of the first and the third is 170
Answers
Answer:
Step-by-step explanation:
Explanation:
Let's say that the first odd integer is
x
. The second consecutive odd integer would have to be
x+2
. (It would not be
x+1
because that would result in an even integer. The sum of any two odd numbers is even.) The third consecutive odd integer would be
(x+2)+2 or x+4.
The sum of the first, twice the second, and three times the third can be written as x+2(x+2)+3(x+4
This equals 70. We can now distribute and solve for x.
x+2(x+2)+3(x+4)=70
x+2x+4+3x+12=70→distributive
6x+16=70→gather like terms
6x=54→subtract 16 from both sides
x=9→divide both sides by 6
Thus, the three consecutive odd integers are 9,11,and 13.
Answer:
As We Know Consecutive Odd Numbers
Comes at Interval of one Even Number
SO ,
Let the No. be : X , X + 2 , X + 4
NOW , A/Q : -
( X ) ^2 + ( X + 4 ) ^ 2 = 170
=> X^2 + X^2 + ( 4 )^2 + 2 { X × 4 } = 170
=> 2X^2 + 16 + 8 X = 170
On Taking 2 As Common
=> 2 [ X^2 + 8 + 4 X ] = 170
=> X^2 + 8 + 4 X = 85
=> X^2 + ( 8 - 85 ) + 4 X = 0
=> X^2 - 77 + 4 X = 0
Or ,
=> X^2 + 4 X - 77 = 0
=> X^2 + 11 X - 7 X - 77 = 0
=> X ( X + 11 ) - 7 ( X + 11 ) = 0
Therefore ,
=> X - 7 = 0 => X = 7
OR ,
=> X + 11 = 0 => X = - 11
Therefore ,
All sets of three consecutive odd integers
Are Following : -
7 , 9 , 11
or ,