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J(n, n+2) k(3n, n-1) jk=5. n=?
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J(n, n+2) k(3n, n-1) jk=5. N=?
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One approach can be to work using following relationship e.g. C^2 = (a2- a1) ^2 + (b2 – b1) ^2…. (1)

Now a1=n, a2=3n & b1= (n=2), b2= (n-1).

Substituting values for a1, a2, b1, b2 in equation (1), it gives:

jk ^ 2 = (3n- n ) ^ 2 + (n-1+n+2) ^2 
5 ^2 = (2n) ^ 2 + (2n +1 )^ 2 
25 = 4n ^2 + 4n ^2 +4n + 1 
0 = 8n ^2 -+4n +1 - 25 

Divide both sides by 4, it gives
8n ^2 + 4n -24 = 0  
2n ^2 + n - 6 = 0 
2n ^2 + 4n - 3n - 6 = 0 
2n (n +2 ) - 3(n+ 2) = 0 
(2n -3) + (n+2) = 0 
n = 3/2 or -2

  
by mathforum
Aug 20, 14 06:47PM PST
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