Question

If x²+y²-6y-10z+z²+38 = -4x . Then find (x+y+z)
please give full solution​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\sf \: {x}^{2} + {y}^{2} - 6y - 10z + {z}^{2} + 38 = - 4x$

$\large\underline{\sf{To\:Find :-}}$

$\sf \: the \: value \: of \: x + y + z$

$\large\underline{\sf{Solution-}}$

Given that

$\sf \: {x}^{2} + {y}^{2} - 6y - 10z + {z}^{2} + 38 = - 4x$

$\sf \: {x}^{2} + {y}^{2} - 6y - 10z + {z}^{2} + 38 + 4x = 0$

Rearrange the terms, we get

$\sf \: ({x}^{2} + 4x) + ( {y}^{2} - 6y) + ({z}^{2} - 10z) + 38 = 0$

$\sf \: ({x}^{2} + 4x + 4 - 4) + ( {y}^{2} - 6y + 9 - 9) + ({z}^{2} - 10z + 25 - 25) + 38 = 0$

$\: \sf \: \: \: \: \: (using \: the \: concept \: of \: completing \: squares)$

$\sf \: {(x + 2)}^{2} - 4+{(y - 3)}^{2}- 9 + {(z - 5)}^{2} - 25 + 38 = 0$

$\bf\implies \: {(x + 2)}^{2} + {(y - 3)}^{2} + {(z - 5)}^{2} = 0$

Since, we know

Sum of squares is 0 only when

$\sf \: x + 2 = 0 \: \: and \: \: y - 3 = 0 \: \: and \: \: z - 5 = 0$

$\bf\implies \:x = - 2 \: \: and \: \: y \: = \: 3 \: \: and \: \: z \: = \: 5$

Hence,

$\rm :\longmapsto\:\sf \: x \: + \: y \: + \: z \: = \: - 2 + 3 + 5 = 6$

Useful Identities :-

$\boxed{ \rm \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }$

$\boxed{ \rm \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }$

$\boxed{ \rm \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}$

$\boxed{ \rm \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}$

$\boxed{ \rm \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} }$

$\boxed{ \rm \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} }$

$\boxed{ \rm \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$