Question

find xy+yz+zx if x+y+z=√42 and x²+y²+z²=16

Answer 1

x+y+z=√42
x²+y²+z²=16

$(x+y+z)^2= x^{2} +y^2+z^2 + 2(xy+yz+zx)\\ \\ \Rightarrow( \sqrt{42} )^2= 16 + 2(xy+yz+zx)\\ \\ \Rightarrow 42 = 16 + 2(xy+yz+zx)\\ \\ \Rightarrow 2(xy+yz+zx)=42-16 = 26\\ \\ \Rightarrow (xy+yz+zx) = \frac{26}{2} \\ \\ \Rightarrow xy+yz+zx=13$