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GrogVix [38]
3 years ago
7

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

Mathematics
2 answers:
alekssr [168]3 years ago
8 0
<em>we know,</em>
<em>(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)</em>
<em>9²  = x² + y² + z² + 2 X 26</em>
<em> 81 = x² + y² + z² + 52       </em>
<em>81 - 52 = x² + y² + z²</em>
<em>therefore, x² + y² + z² = 29</em>
vichka [17]3 years ago
7 0
<em>First, we must expand  and simplify :</em>
x + y+z = 9

(x + y+z)^{2} = (9)^{2}

x^{2} + y^{2}+z^{2} + 2(xy + yz+zx) = 81

<em>And Than, Enter the value :</em>

x^{2} + y^{2}+z^{2} + 2(26) = 81

x^{2} + y^{2}+z^{2} + 52 = 81

x^{2} + y^{2}+z^{2} = 81- 52

\boxed{x^{2} + y^{2}+z^{2} = 29}

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