Question

If x+y+z=7 and x^2+y^2+z^2=19,then find xy+yz+zx

Answer 1

Hiii. .....friend

Here is ur answer,

=> x+y+z =7 .

$= > \: {x}^{2} + {y}^{2} + {z}^{2} = 19$

We know that,

$= > \: ( {x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx$

$= > \: {7}^{2} = 19 + 2(xy + yz + zx)$

$= > \: 2(xy + yz + zx) = 49 - 19 = 30$

xy+yz+zx => 30/2.

=> 15.


:-)Hope it helps u.