Dooravaninagar, Bangalore, India - 560016.
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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 (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
17
Board
CBSE, ICSE, State
CBSE Subjects taught
Hindi, Mathematics, Science, English, Computers
ICSE Subjects taught
English, Science, Mathematics, Hindi, Computer science
Taught in School or College
No
State Syllabus Subjects taught
Computer Science, Mathematics, Hindi, English, Science
1. Which school boards of Class 1-5 do you teach for?
CBSE, ICSE and State
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 Learn Tuition
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 (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
17
Board
CBSE, ICSE, State
CBSE Subjects taught
Hindi, Mathematics, Science, English, Computers
ICSE Subjects taught
English, Science, Mathematics, Hindi, Computer science
Taught in School or College
No
State Syllabus Subjects taught
Computer Science, Mathematics, Hindi, English, Science
Answered on 24/06/2020 Learn Tuition
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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