(2x −3y +2z +4)^(2 ) +(x−y−z)^2 +(x+y+z−8)^2 =0 find x , y, z


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Question Number 187069 by Humble last updated on 13/Feb/23

${(2 x - 3 y + 2 z + 4)}^{2} + {(x - y - z)}^{2} + {(x + y + z - 8)}^{2} = 0$ $f i n d$ $x, y, z$
	Answered by ARUNG_Brandon_MBU last updated on 13/Feb/23
	$(2x−3y+2z+4)^2 +(x−y−z)^2 +(x+y+z−8)^2 =0 { ((2x−3y+2z+4=0)),((x−y−z=0)),((x+y+z−8=0)) :}⇒ { ((2x−3y+2z=−4),(...(i))),((x−y−z=0),(...(ii))),((x+y+z=8),(...(iii))) :} (ii)+(iii) ⇒x=4 x=4 in (i) and (iii) ⇒ { ((3y−2z=12)),((y+z=4)) :} ⇒y=4, z=0, x=4$
	${(2 x - 3 y + 2 z + 4)}^{2} + {(x - y - z)}^{2} + {(x + y + z - 8)}^{2} = 0$ ${\begin{cases} 2 x - 3 y + 2 z + 4 = 0 \\ x - y - z = 0 \\ x + y + z - 8 = 0 \end{cases} \Rightarrow {\begin{cases} 2 x - 3 y + 2 z = - 4 & . . . (i) \\ x - y - z = 0 & . . . (i i) \\ x + y + z = 8 & . . . (i i i) \end{cases}$ $(i i) + (i i i) \Rightarrow x = 4$ $x = 4 in (i) and (i i i)$ $\Rightarrow {\begin{cases} 3 y - 2 z = 12 \\ y + z = 4 \end{cases} \Rightarrow y = 4, z = 0, x = 4$
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