Question 255744: The sum of the squares of two positive consecutive integers is 41. Find integers. Show steps please^^ Found 3 solutions by drk, checkley77, CharlesG2:Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! Let x = first integer and x+1 = consecutive integer.
(i)
step 1 - expand (i) to get
(ii)
step 2 - set (i) =0 and factor as
(iii)
divide by 2 to get
x-4 = 0 - - -> x = 4
x+5=0 - - - > x = -5
SInce we want positive, we have
x = 4 and x+1 = 5.
You can put this solution on YOUR website! x & x+1 are the 2 consecutive numbers.
x^2+(x+1)=41
x^2+x^2+2x+1=41
2x^2+2x+1-41=0
2x^2+2x-40=0
2(x^2+x-20)=0
2(x+5)(x-4)=0
x-4=0
x=4 ans.
4+1=5 ans.
Proof:
4^2+5^2=41
16+25=41
41=41
You can put this solution on YOUR website! The sum of the squares of two positive consecutive integers is 41. Find integers. Show steps please^^
integer I
next consecutive integer I+1
I^2+(I+1)^2=41
I^2+((I+1)(I+1))=41
I^2+I^2+2I+1=41
2I^2+2I+1=41
2I^2+2I-40=0
I^2+I-20=0
(I+5)(I-4)=0
I=-5 and I=4
-5 is not positive
so I is 4
I+1=5
integers are 4 and 5
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