If x + y + z = 5 and x^2 + y^2 + z^2 = 13 then xy + yz + zx = ?
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Answer:
Given,
x−y+z=5 and x
2
+y
2
+z
2
=49
We know,
(a+b+c)
2
=a
2
+b
2
+c
2
+2(ab+bc+ca)
Here,
a=x
b=−y
c=z
Thus,
=>(x−y+z)
2
=x
2
+y
2
+z
2
+2(−xy−yz+zx)
=>5
2
=49+2(−xy−yz+zx)
=>−2(−xy−yz+zx)=49−25
=>(−xy−yz+zx)=−
2
24
=>(−xy−yz+zx)=−12
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