Mathematics

if x+y+z=9, xy+yz+zx =26; find x²+y²+z²


Answers

TamaraTewksbury170

4 years ago Comment

First, we must expand  and simplify :
[latex]x + y+z = 9[/latex]

[latex](x + y+z)^{2} = (9)^{2}[/latex]

[latex]x^{2} + y^{2}+z^{2} + 2(xy + yz+zx) = 81[/latex]

And Than, Enter the value :
[latex]x^{2} + y^{2}+z^{2} + 2(26) = 81[/latex]

[latex]x^{2} + y^{2}+z^{2} + 52 = 81[/latex]

[latex]x^{2} + y^{2}+z^{2} = 81- 52[/latex]

[latex]\boxed{x^{2} + y^{2}+z^{2} = 29}[/latex]

Malafronte41

4 years ago Comment

we know,
(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)
9²  = x² + y² + z² + 2 X 26
 81 = x² + y² + z² + 52       
81 - 52 = x² + y² + z²
therefore, x² + y² + z² = 29