a + b + c = 10 and a^2 + b^2 + c^2 = 38, find ab + bc + ca

Using the formula, [∵ (x \+ y \+ z) 2 = x 2 \+ y 2 \+ z 2 \+ 2xy \+ 2yz \+ 2zx] So, ⇒ (a \+ b \+ c) 2 = a 2 \+ b 2 \+ c 2 \+ 2ab \+ 2bc \+ 2ca ⇒ (a \+ b \+ c) 2 = a 2 \+ b 2 \+ c 2 \+ 2(ab \+ bc \+ ca) Putting (a \+ b \+ c) = 10 and (a 2 \+ b 2 \+ c 2 ) = 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 = ⇒ ab \+ bc \+ ca = 31 Hence, the value of (ab \+ bc \+ ca) is 31.

Mathematics

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

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Mathematics

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

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

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

Find : a² + b² + c², if a + b + c = 9 and ab + bc + ca = 24.

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