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Question
If a + b + c = 10 and a2 + b2 + c2 = 38, find : ab + bc + ca
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Solution
a + b + c = 10
⇒ (a + b + c)2 = (10)2
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 100
⇒ 38 + 2(ab + bc + ca) = 100
⇒ 2(ab + bc + ca) = 100 − 38
⇒ 2(ab + bc + ca) = 62
⇒ (ab + bc+ ca) =`62/2`
⇒ ab + bc + ca = 31
Alternative Method :
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (10)2 = 38 + 2(ab + bc + ca)
⇒ 100 = 38 + 2(ab + bc + ca)
⇒ 100 − 38 = 2(ab + bc + ca)
⇒ 62 = 2(ab + bc + ca)
⇒ `62/2` = ab + bc + ca
⇒ 31 = ab + bc + ca
∴ ab + bc + ca = 31
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