Hoskote, Hoskote Details verified of Sree Sujana K.✕
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Sri Krishnadevaraya University 2008
Master of Science (M.Sc.)
Hoskote, Hoskote, India - 562114
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
5
Board
State, CBSE
Preferred class strength
One on one/ Private Tutions
Subjects taught
Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
5
Board
State, CBSE
Preferred class strength
One on one/ Private Tutions
Subjects taught
Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 8 Tuition
5
Board
CBSE, State, ICSE
Experience in School or College
• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.
Subjects taught
Science, Physics, Mathematics, Chemistry
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
6
Board
State, CBSE, ICSE
Experience in School or College
• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.
Subjects taught
Physics, EVS, Science, Chemistry, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
6
Board
State, CBSE
Experience in School or College
• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.
Subjects taught
Mathematics
Taught in School or College
Yes
Answered on 26/06/2019 Learn CBSE - Class 10/Mathematics
Ask a Question
Post a Lesson
Given, f(x) = x4 + 4x2 + 5
= (x2)2 + 4x2 + 5
Put x2 = a
Hence x4 + 4x2 + 5 = (x2)2 + 4x2 + 5
= a2 + 4a + 5
To find the zero of the above polynomial, we take a2 + 4a + 5 = 0
The above equation does not have a real root
Hence there is no zero for this given polynomial.
Like 0 
Answers 13 
Comments Answered on 26/06/2019
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Post a Lesson
RHS= (x+y+z)(x2+y2+z2-xy-yz-zx)
=x(x2+y2+z2-xy-yz-xz)+y(x2+y2+z2-xy-yz-zx)+z(x2+y2+z2-xy-yz-zx)
=x3+xy2+xz2-x2y-xyz+x2y+y3+z2y-y2x-y2z-xyz+x2z+y2z+z3-xyz-yz2-z2x
=x3+y3+z3+(x2y-x2y)+(y2z-y2z)+(y2z-y2z)-3xyz
=x3+y3+z3-3xyz
=LHS
Hence proved.
Like 1 Answers 278 Comments Answered on 26/06/2019 Learn CBSE - Class 10/Mathematics
Ask a Question
Post a Lesson
Like 0 Answers 13 Comments Answered on 18/06/2019
Ask a Question
Post a Lesson
Like 0 Answers 986 Comments Ask a Question
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
5
Board
State, CBSE
Preferred class strength
One on one/ Private Tutions
Subjects taught
Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
5
Board
State, CBSE
Preferred class strength
One on one/ Private Tutions
Subjects taught
Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 8 Tuition
5
Board
CBSE, State, ICSE
Experience in School or College
• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.
Subjects taught
Science, Physics, Mathematics, Chemistry
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
6
Board
State, CBSE, ICSE
Experience in School or College
• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.
Subjects taught
Physics, EVS, Science, Chemistry, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
6
Board
State, CBSE
Experience in School or College
• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.
Subjects taught
Mathematics
Taught in School or College
Yes
Answered on 26/06/2019 Learn CBSE - Class 10/Mathematics
Ask a Question
Post a Lesson
Given, f(x) = x4 + 4x2 + 5
= (x2)2 + 4x2 + 5
Put x2 = a
Hence x4 + 4x2 + 5 = (x2)2 + 4x2 + 5
= a2 + 4a + 5
To find the zero of the above polynomial, we take a2 + 4a + 5 = 0
The above equation does not have a real root
Hence there is no zero for this given polynomial.
Like 0 Answers 13 Comments Answered on 26/06/2019
Ask a Question
Post a Lesson
RHS= (x+y+z)(x2+y2+z2-xy-yz-zx)
=x(x2+y2+z2-xy-yz-xz)+y(x2+y2+z2-xy-yz-zx)+z(x2+y2+z2-xy-yz-zx)
=x3+xy2+xz2-x2y-xyz+x2y+y3+z2y-y2x-y2z-xyz+x2z+y2z+z3-xyz-yz2-z2x
=x3+y3+z3+(x2y-x2y)+(y2z-y2z)+(y2z-y2z)-3xyz
=x3+y3+z3-3xyz
=LHS
Hence proved.
Like 1 Answers 278 Comments Answered on 26/06/2019 Learn CBSE - Class 10/Mathematics
Ask a Question
Post a Lesson
Like 0 Answers 13 Comments Answered on 18/06/2019
Ask a Question
Post a Lesson
Like 0 Answers 986 Comments Ask a Question
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