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Mathematics

a + b + c = 10 and a2 + b2 + c2 = 38, find ab + bc + ca

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Answer

Using the formula,

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

So,

⇒ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

Putting (a + b + c) = 10 and (a2 + b2 + c2) = 38, we get

⇒ (10)2 = 38 + 2(ab + bc + ca)

⇒ 100 = 38 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 100 - 38

⇒ 2(ab + bc + ca) = 62

⇒ ab + bc + ca = 622\dfrac{62}{2}

⇒ ab + bc + ca = 31

Hence, the value of (ab + bc + ca) is 31.

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