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

Sagot :

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

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

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

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

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

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

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