Question

if x+y+z=4and find the value of (xy+yz+zx)and x2+y2+z2=10​

Answer 1

SOLUTION :

$\sf(x + y + z) = 4$

$\sf \implies {(x + y + z)}^{2} = {4}^{2}$

$\sf \implies \: {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx) = 16$

Putting value,

$\sf \implies \: 10 + 2(xy + yz + zx) = 16$

$\sf \implies {\boxed{ \mathfrak{ (xy + yz + zx) = 3}}}$

Answer 2

Answer:

(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−xz)

=>4

2

=50+2(xy−yz−xz)

=>

2

16−50

=(xy−yz−xz)

=>(xy−yz−xz)=−17