# 90) If ${x}^{2}+4{y}^{2}+9{z}^{2}-2x-12y+10=0$ then the value of x + y + z is (1) 2.5 (2) 2.4 (3) 2.3 (4) 2.0

Dear Student,

${x}^{2}+4{y}^{2}+9{z}^{2}-2x-12y+10=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x+1+4{y}^{2}-12y+9+9{z}^{2}=0\phantom{\rule{0ex}{0ex}}⇒\left({x}^{2}-2x+{1}^{2}\right)+\left({\left(2y\right)}^{2}-12y+{3}^{2}\right)+{\left(3z\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(2y-3\right)}^{2}+{\left(3z\right)}^{2}=0$
As we know that square of any number is positive. Sum of three positive numbers is zero only if all the three numbers are zero.

Regards,

• 4
X+ 4Y+ 9Z- 2X- 12Y+10 = 0
(X-1)2+(2Y-3)2+(3Z)2=0
AS ALL VALUE ARE NON-NEGATIVE BECAUSE OF SQUARE, therefore ADDITION OF NON-NEGATIVE NO. IS ZERO WHEN ALL THE VALUES ARE ZERO.
SO,  X=1 ,Y=3/2 & Z=0
HENCE X+Y+Z= 1+3/2+0=5/2=2.5
• 0
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