if x,y,z E r and 121x²+4y²+9z²-22x+4y+6z+3=0
Answers
Answered by
0
Answer:
to see the answer please see the attached attachment and mark me as brainilist
Attachments:
Answered by
2
Answer:
if x,y,z E r and 121x²+4y²+9z²-22x+4y+6z+3=0
[(11x)^2-22x+1] + [(2y)^2+4y+1]+[(3z)^2+6z+1]=0
(11x-1)^2+(2y+1)^2+(3z+1)^2
We know that squares of real no. are always positive therefore sum of their square can't be 0 unless and until they are zero
x=1/11
y=-1/2
z=-1/3
Step-by-step explanation:
Similar questions