Question

find value of x + Y + Z if X square + Y square + Z square equal to 18 and xy + Y Z plus ZX is equals to 9​

Answer 1

Answer:

$\green{x+y+z=6}$

Step-by-step explanation:

$\boxed{Given}$

$x {}^{2} + y {}^{2} + z {}^{2} = 18$

$xy + yz + zx = 9$

$\boxed{Tofind}$

value of

$x + y + z$

$\boxed{Answer}$

using identity

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

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

now putting the given values, we get

$(x + y + z) {}^{2} = 18 + 2(9)$

$(x + y + z) {}^{2} = 18 + 18$

$(x + y + z) {}^{2} = 36$

$x + y + z = \sqrt{36}$

$x + y + z = 6$

Answer 2

Answer:

Answer:

\green{x+y+z=6}x+y+z=6

Step-by-step explanation:

\boxed{Given}

Given

x {}^{2} + y {}^{2} + z {}^{2} = 18x

2

+y

2

+z

2

=18

xy + yz + zx = 9xy+yz+zx=9

\boxed{Tofind}

Tofind

value of

x + y + zx+y+z

\boxed{Answer}

Answer

using identity

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

2

=x

2

+y

2

+z

2

+2xy+2yz+2zx

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

2

+y

2

+z

2

+2(xy+yz+zx)

now putting the given values, we get

(x + y + z) {}^{2} = 18 + 2(9)(x+y+z)

2

=18+2(9)

(x + y + z) {}^{2} = 18 + 18(x+y+z)

2

=18+18

(x + y + z) {}^{2} = 36(x+y+z)

2

=36

x + y + z = \sqrt{36}x+y+z=

36

x + y + z = 6x+y+z=6