hey frnds! plz help me find two consecutive positive integers sum of whose squares is 365
Answers
Answered by
1
let the numbers be x and x+1
x^2+(x+1)^2=365
x^2+x^2+2x+1=365
2x^2+2x+1=365
2x^2+2x-364=0
2(x^2+x-182)=0
x^2+x-182=0
x^2+14x-13x-182=0
x(x+14)-13(x+14)=0
(x-13)(x+14)=0
so, x=13 or x=-14
as x cannot be negative according to the question, x=13
so the consecutive numbers are 13 and 14
sorry for the delay
x^2+(x+1)^2=365
x^2+x^2+2x+1=365
2x^2+2x+1=365
2x^2+2x-364=0
2(x^2+x-182)=0
x^2+x-182=0
x^2+14x-13x-182=0
x(x+14)-13(x+14)=0
(x-13)(x+14)=0
so, x=13 or x=-14
as x cannot be negative according to the question, x=13
so the consecutive numbers are 13 and 14
sorry for the delay
Anonymous:
what is your name
sorry can't say
why
OK fyn
bye
nice to meet you
my pleasure to meet you
bye
thanks
welcome dear
Answered by
2
Hi gd evening....
dear...
will u please Inbox me...
please please please...
I really want your help...
please....
dear
please....
please...
accept my request please..
Similar questions