Question

If x+y+z=9 and xy + yz + zx = 23 then find the value of (x² + y² +z²).​

Answer 1

The value of ($x^{2} +y^{2} + z^{2}$) is 35.

• Given that x+y+z = 9 and xy+yz+zx = 23.
• We have to find the value of ($x^{2} +y^{2} + z^{2}$).
• Now we first take the given value , x+y+z = 9

Squaring both sides , we get

$(x+y+z)^2$ = $9^2$

Expanding the expression $(x+y+z)^2$ ,

$x^{2} +y^{2} +z^{2} + 2xy+2yz+2zx$ = 81

• Now it is given that ,  xy+yz+zx = 23

$x^{2} +y^{2} +z^{2} + 2( xy + yz + zx )$ = 81

$x^{2} +y^{2} +z^{2} + 2( 23 )$ = 81

$x^{2} +y^{2} +z^{2} + 46$ = 81

$x^{2} +y^{2} +z^{2}$ = 81 - 46

$x^{2} +y^{2} +z^{2}$ = 35

Answer 2

Step-by-step explanation:

see this attachment.....