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J(n, n+2) k(3n, n-1) jk=5. n=?
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by
Ja
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J(n, n+2) k(3n, n-1) jk=5. N=?
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Responses
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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 |
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by
mathforum
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Aug 20, 14 06:47PM PST
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