if x+ y+ z=0 x2 +y2+4z=0 xy+yz+zx=?
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Given:
x+ y+ z=0
and
x2 +y2+4z=0
Using identity (a+b+c)2 = a2 + b2 + c2 + 2ab+ 2bc + 2ca
similarly,
(x+y+z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= x2 + y2 + z2 + 2 (xy + yz + zx)
substitute for x + y + z and x2 + y2 + x2
and you'll get
(0)^2 = x2 + y2 + z2 + 2 (xy + yz + zx)
0 = 0 + 2 (xy+ yz + zx)
0 = 2 (xy + yz + zx)
0 = xy+ yz + zx
so,
xy + yz + zx =0
x+ y+ z=0
and
x2 +y2+4z=0
Using identity (a+b+c)2 = a2 + b2 + c2 + 2ab+ 2bc + 2ca
similarly,
(x+y+z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= x2 + y2 + z2 + 2 (xy + yz + zx)
substitute for x + y + z and x2 + y2 + x2
and you'll get
(0)^2 = x2 + y2 + z2 + 2 (xy + yz + zx)
0 = 0 + 2 (xy+ yz + zx)
0 = 2 (xy + yz + zx)
0 = xy+ yz + zx
so,
xy + yz + zx =0
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