UPSC Exam > UPSC Questions > Let m and n be the roots of the equation ax²+...
Start Learning for Free
Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²?
Most Upvoted Answer
Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b...
Given Information:
- Roots of the quadratic equation: m, n
- Coefficients of the quadratic equation: a, b, c in a non-constant GP
- Sum of reciprocals of roots: 1/m + 1/n = 1/2
Analysis:
- Let the roots be in the form m = 1/x and n = 1/(x+r), where r is a constant
- From the sum of reciprocals of roots, we get: 1/m + 1/n = x(x+r)/x(x+r) = (2x+r)/(x^2 + xr) = 1/2
- Solving the above equation, we get: r = x
- Therefore, m = 1/x and n = 1/(2x)
Calculation:
- From the quadratic equation, we have: m + n = -b/a = -a^2/b
- Substituting the values of m and n, we get: 1/x + 1/(2x) = -a^2/b
- Simplifying the above equation, we get: 3x = -a^2/b
- Therefore, a = -3x/b
Final Calculation:
- Substituting the value of a in the sum of reciprocals equation, we get: (2x + 3x)/(9x^2) = 1/2
- Solving the above equation, we get: x = 1/3
- Therefore, m = 3, n = 6
- Hence, (m-n)^2 = (3-6)^2 = 9
- Roots of the quadratic equation: m, n
- Coefficients of the quadratic equation: a, b, c in a non-constant GP
- Sum of reciprocals of roots: 1/m + 1/n = 1/2
Analysis:
- Let the roots be in the form m = 1/x and n = 1/(x+r), where r is a constant
- From the sum of reciprocals of roots, we get: 1/m + 1/n = x(x+r)/x(x+r) = (2x+r)/(x^2 + xr) = 1/2
- Solving the above equation, we get: r = x
- Therefore, m = 1/x and n = 1/(2x)
Calculation:
- From the quadratic equation, we have: m + n = -b/a = -a^2/b
- Substituting the values of m and n, we get: 1/x + 1/(2x) = -a^2/b
- Simplifying the above equation, we get: 3x = -a^2/b
- Therefore, a = -3x/b
Final Calculation:
- Substituting the value of a in the sum of reciprocals equation, we get: (2x + 3x)/(9x^2) = 1/2
- Solving the above equation, we get: x = 1/3
- Therefore, m = 3, n = 6
- Hence, (m-n)^2 = (3-6)^2 = 9
Attention UPSC Students!
To make sure you are not studying endlessly, EduRev has designed UPSC study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in UPSC.
|
Explore Courses for UPSC exam
|
|
Similar UPSC Doubts
Top Courses for UPSCView all
Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²?
Question Description
Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²?.
Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²?.
Solutions for Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? in English & in Hindi are available as part of our courses for UPSC.
Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free.
Here you can find the meaning of Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? defined & explained in the simplest way possible. Besides giving the explanation of
Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²?, a detailed solution for Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? has been provided alongside types of Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? theory, EduRev gives you an
ample number of questions to practice Let m and n be the roots of the equation ax²+bx-c=0, where a≠0. If a,b and c are the consecutive terms of a non constant GP and 1/m+ 1/n =1/2, then find the value of (m-n)²? tests, examples and also practice UPSC tests.
|
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup