Question
If (x + y + z) = 10 and (x 2 + y 2
+ z 2 ) = 52, then find {z(x + y) + xy}.Solution
Since, (x + y + z) 2 = x 2 + y 2 + z 2 + 2(xy +yz +zx)
So, (10) 2 = 52 + 2(xy +yz +zx)
Or, 100 - 52 = 2(xy +yz +zx)
Or, 48 = 2(xy +yz +zx)
Or, 24 = (xy +yz +zx)
So, z(x + y) + xy = 24
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