find two consecutive positive integers, sum of whose square is 365.
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Hey there,
Solution :
Let the consecutive positive integers be xand x + 1.
Given that x2 +(x+1)2 = 365
= x2 + x2 +1+2x = 365
= 2x2 +2x – 364 = 0
= x2 + x – 182 = 0
= x2 + 14x – 13x – 182 = 0
= x(x+14) – 13(x+14) = 0
= (x+14)(x – 13) = 0
Either x + 14 = 0 or x − 13 = 0, i.e., x = −14 or x= 13
Since, the integers are positive, x can be 13.
so,x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers=13 and 14.
:-) Be happy
Solution :
Let the consecutive positive integers be xand x + 1.
Given that x2 +(x+1)2 = 365
= x2 + x2 +1+2x = 365
= 2x2 +2x – 364 = 0
= x2 + x – 182 = 0
= x2 + 14x – 13x – 182 = 0
= x(x+14) – 13(x+14) = 0
= (x+14)(x – 13) = 0
Either x + 14 = 0 or x − 13 = 0, i.e., x = −14 or x= 13
Since, the integers are positive, x can be 13.
so,x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers=13 and 14.
:-) Be happy
johntuscumbia:
it's is from internet
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