Question

the value of m+n+P + z = If x^2 + y^2 + z^2 = 78 and xy + yz + zx = 59, then find the value of x+y+z. ​

Answer 1

we know that $(x+y+z)^{2} =x^{2}+ y^{2}+ z^{2} +2*(xy+yz+zx)$

Step-by-step explanation:

• [tex](x+y+z)^{2} =x^{2}+ y^{2}+ z^{2} +2*(xy+yz+zx)\\\\ (x+y+z)^{2}=78+2*59\\=78+118=196\\\\ (x+y+z)^{2}=196\\\\ (x+y+z)=\sqrt{196} =14 [/tex]

the value of x+y+z=14