2006-09-17 19:43:44 · 15 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
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2006-09-17 20:00:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋
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2006-09-17 23:11:26 · answer #2 · answered by c2 brahmin 2 · 0⤊ 0⤋
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2006-09-17 19:51:59 · answer #3 · answered by Sweety Tweety 2 · 0⤊ 0⤋
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2006-09-17 19:47:51 · answer #4 · answered by Demiurge42 7 · 0⤊ 0⤋
We know that, in general:
a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca).
Consequently, substituting the given values, we obtain:
a^2 + b^2 + c^2
= 5 - 2*2
= 1.
2006-09-17 21:34:27 · answer #5 · answered by K Sengupta 4 · 0⤊ 0⤋
(a+b+c)^2 = a^2+b^2+c^2 +2(ab+bc+ca)
putting the values
5 = a^2+b^2+c^2+2(2)
a^2+b^2+c^2 = 5 - 2* 2= 1
2006-09-17 19:54:48 · answer #6 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
[a^2 + b^2 +c^2 = (a+b+c+)^2 - 2(ab+bc+ca)]
put up values!
a^2 +b^2 +c^2 = (5) -2(2)
so,
a^2 +b^2 +c^2 =1 is the answer!
2006-09-17 21:17:04 · answer #7 · answered by testy c 1 · 0⤊ 0⤋
Let xa = (a+b+c) = 0 --- --- --- --- --- --- --- --- (1); ya = xa^2 = (a+b+c)^2 = a^2 +b^2 + c^2 +2ab + 2bc + 2ca --- --- --- --- --- --- --- --- (2); ya = 0; So, from (2), a^2 +b^2 + c^2 = -2(ab + bc + ca) --- --- --- --- --- --- --- --- (3); Let, p = a^2 +b^2 + c^2, --- --- --- --- --- --- --- --- (4); and q = -(ab + bc + ca) --- --- --- --- --- --- --- --- (5); From (3), p/q = 2 --- --- --- --- --- --- --- --- --- --- --- --- --- (6); Now, let us dissect the LHS with its terms; From (1), (a+b+c) = 0 --- --- --- --- --- --- --- --- (1); Multiply eqn (1) by a; We get, a(a+b+c) = 0 that is, a^2 + ab + ca = 0 that is, a^2 - bc = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (7); Similarly, multiplying eqn (1) by b, we can get, b^2 - ca = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (8); And again, multiplying eqn (1) by c, we can get, c^2 - ab = -(ab + bc+ ca) --- --- --- --- --- --- --- --- (9); Thus, using (7), (8) & (9), we can rearrange the LHS as follows; LHS = a^2/(a^2-bc) + b^2/(b^2-ca) + c^2/(c^2-ab) = (a^2 +b^2+ c^2)/(-ab-bc-ca) --- --- --- --- (10); From (4), (5) & (10), we get, LHS = p/q --- --- --- --- --- --- --- --- (11); And, from (6) & (11), we get, LHS = 2 that is, a^2/(a^2-bc) + b^2/(b^2-ca) + c^2/(c^2-ab) = 2 --- --- --- --- --- --- --- --- (proved);
2016-03-27 06:46:47 · answer #8 · answered by Anonymous · 0⤊ 0⤋
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
So
a^2+b^2+c^2= (a+b+c)^2 - 2(ab+bc+ca)
= 5 - 2(2)
= 5-4
= 1
2006-09-17 19:55:13 · answer #9 · answered by Mank 3 · 0⤊ 0⤋
(a+b+c)^2 = 5
a^2+ab+ac+ba+b^2+bc+ca+cb+c^2 = 5
a^2+b^2+c^2+2ab+2bc+2ca = 5
a^2+b^2+c^2+2(ab+bc+ca) = 5
a^2+b^2+c^2 + 2(2) = 5 substituting ab+bc+ca=2
a^2+b^2+c^2+4 = 5
a^2+b^2+c^2 = 5-4
a^2+b^2+c^2 = 1
2006-09-17 22:29:33 · answer #10 · answered by uselessadvice 4 · 0⤊ 0⤋