If $x + y + z = 10$ and $x^{2} + y^{2} + z^{2} = 40$ find $x y + y z + z x$

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Solution

It is given that $x + y + z = 10$ and $x^{2} + y^{2} + z^{2} = 40$ .

Consider the square of $(x + y + z)^{2}$ as shown below:

$(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x (∵ (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a) \Rightarrow (10)^{2} = 40 + 2 (x y + y z + z x) (G i v e n x + y + z = 10, x^{2} + y^{2} + z^{2} = 40) \Rightarrow 100 = 40 + 2 (x y + y z + z x) \Rightarrow 100 - 40 = 2 (x y + y z + z x) \Rightarrow 2 (x y + y z + z x) = 60 \Rightarrow x y + y z + z x = \frac{60}{2} \Rightarrow x y + y z + z x = 30$

Hence, $x y + y z + z x = 30$ .

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