Dooravaninagar, Bangalore, India - 560016.
Verified 17 yrs of Exp
Details verified of Kapil Garg✕
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Hindi Mother Tongue (Native)
English Proficient
Punjabi Basic
Punjab Engineering College 2000
Bachelor of Engineering (B.E.)
Dooravaninagar, Bangalore, India - 560016
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Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
17
Board
CBSE, State, ICSE
Subjects taught
English, Science, Computers, Hindi, Mathematics, Computer Science
Taught in School or College
No
Upcoming Live Classes
1. Which school boards of Class 1-5 do you teach for?
CBSE, State and ICSE
2. Have you ever taught in any School or College?
No
3. Which classes do you teach?
I teach Class I-V Tuition Class.
4. Do you provide a demo class?
Yes, I provide a free demo class.
5. How many years of experience do you have?
I have been teaching for 17 years.
Answered on 24/06/2020
5(X^2+Y^2+Z^2)=4(XY+YZ+ZX).
(X^2+Y^2+Z^2) is always positive since it is a sum of squares.
=> left-hand side term is still positive, so by equality; the right-hand side term is also positive
:: The first conclusion:- XY+YZ+ZX is always positive.
Now it can be re-written as
4(X^2+Y^2+Z^2) + (X^2+Y^2+Z^2) = 4(XY+YZ+ZX)
=> 4(X^2+Y^2+Z^2) - 4(XY+YZ+ZX) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0
=> 4((X^2+Y^2+Z^2) - (XY+YZ+ZX)) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0
assuming (X^2+Y^2+Z^2) to be a positiver term A and
(XY+YZ+ZX) to be a positive term B
=> 4(A - B) + A = 0
=> the above relation can only be true when both positive terms A & B are 0.
=> (X^2+Y^2+Z^2) = 0
=> the above can only be true when each of X, Y & Z are 0, since squares of each will always be positive.
Thus X = Y = Z = 0 => X:Y:Z = 1:1:1
Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
17
Board
CBSE, State, ICSE
Subjects taught
English, Science, Computers, Hindi, Mathematics, Computer Science
Taught in School or College
No
Answered on 24/06/2020
5(X^2+Y^2+Z^2)=4(XY+YZ+ZX).
(X^2+Y^2+Z^2) is always positive since it is a sum of squares.
=> left-hand side term is still positive, so by equality; the right-hand side term is also positive
:: The first conclusion:- XY+YZ+ZX is always positive.
Now it can be re-written as
4(X^2+Y^2+Z^2) + (X^2+Y^2+Z^2) = 4(XY+YZ+ZX)
=> 4(X^2+Y^2+Z^2) - 4(XY+YZ+ZX) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0
=> 4((X^2+Y^2+Z^2) - (XY+YZ+ZX)) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0
assuming (X^2+Y^2+Z^2) to be a positiver term A and
(XY+YZ+ZX) to be a positive term B
=> 4(A - B) + A = 0
=> the above relation can only be true when both positive terms A & B are 0.
=> (X^2+Y^2+Z^2) = 0
=> the above can only be true when each of X, Y & Z are 0, since squares of each will always be positive.
Thus X = Y = Z = 0 => X:Y:Z = 1:1:1
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Dooravaninagar, Bangalore 
