Question

if x+y+z=12and x2+y2+z2=62 then find xy+yz+zx.

Answer 1

Here's your answer !!

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$= x + y + z = 12$

By squaring both side ,we get :-

${(x + y + z)}^{2} = {(12)}^{2}$

${x}^{2} + {y}^{2}+{z}^{2} + 2xy + 2yz + 2zx = 144$

$( {x}^{2} + {y}^{2} + {z}^{2} ) + 2(xy + yz + zx )= 144$

$given= {x}^{2} + {y}^{2} + {z}^{2} = 62$

$= 62 + 2(xy + yz + zx) = 144$

$= 2(xy + yz + zx) = 144 - 62$

$= 2(xy + yz + zx) = 82$

$= xy + yz + zx = \frac{82}{2}$

$= > xy + yz + zx = 41$

Hence,

$= > xy + yz + zx = 41$
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Hope it helps you! :)

Answer 2

Given:- x+y+z=12 and x²+y²+z²=62

then to find:- xy+yz+zx,

x+y+z=12

Squaring both sides,

(x+y+z)² = (12)²

⇒x²+y²+z²+2xy+2yz+2zx = 144

⇒ 62 + 2(xy+yz+zx) = 144

⇒ 2(xy+yz+zx) = 144-62

⇒ xy+yz+zx = 82/2

⇒ xy+yz+zx = 41

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