swethasabu
swethasabu
10-06-2014
Mathematics
contestada
if x+y+z=9, xy+yz+zx =26; find x²+y²+z²
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Icuk18sugiarto
Icuk18sugiarto
10-06-2014
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]
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Аноним
Аноним
10-06-2014
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
Answer Link
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