JEE Exam > JEE Questions > If a, b, c are real, then both the roots of t...
Start Learning for Free
If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are always
- a)positive
- b)negative
- c)real
- d)imaginary
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
If a, b, c are real, then both the roots of the equation (x – b)...
Most Upvoted Answer
If a, b, c are real, then both the roots of the equation (x – b)...
We need to prove that both the roots of the given equation are positive.
Let's first find the discriminant of the given equation:
D = b^2 - 4ac
= (a + b + c)^2 - 4ac
= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4ac
= a^2 + b^2 + c^2 - 2ac + 2ab + 2bc
= (a - c)^2 + b^2 + 2ab + 2bc
Since a, b, and c are real, we know that D ≥ 0 for the given equation to have real roots.
Now, let's find the roots using the quadratic formula:
x = (-b ± √D) / 2a
= (-(a + b + c) ± √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Let's consider the positive root:
x = (-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Since D ≥ 0, we know that the term inside the square root is non-negative. Therefore,
(a - c)^2 + b^2 + 2ab + 2bc ≥ 0
a^2 - 2ac + c^2 + b^2 + 2ab + 2bc ≥ 0
(a + b)^2 + (b + c)^2 + (c - a)^2 ≥ 0
This inequality holds true for all real values of a, b, and c. Therefore,
(-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a > 0
x > 0
Similarly, we can prove that the negative root is also positive:
x = (-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
(-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a < 0="" />
x < 0="" />
Therefore, both roots of the given equation are positive.
Let's first find the discriminant of the given equation:
D = b^2 - 4ac
= (a + b + c)^2 - 4ac
= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4ac
= a^2 + b^2 + c^2 - 2ac + 2ab + 2bc
= (a - c)^2 + b^2 + 2ab + 2bc
Since a, b, and c are real, we know that D ≥ 0 for the given equation to have real roots.
Now, let's find the roots using the quadratic formula:
x = (-b ± √D) / 2a
= (-(a + b + c) ± √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Let's consider the positive root:
x = (-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Since D ≥ 0, we know that the term inside the square root is non-negative. Therefore,
(a - c)^2 + b^2 + 2ab + 2bc ≥ 0
a^2 - 2ac + c^2 + b^2 + 2ab + 2bc ≥ 0
(a + b)^2 + (b + c)^2 + (c - a)^2 ≥ 0
This inequality holds true for all real values of a, b, and c. Therefore,
(-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a > 0
x > 0
Similarly, we can prove that the negative root is also positive:
x = (-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
(-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a < 0="" />
x < 0="" />
Therefore, both roots of the given equation are positive.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer?
Question Description
If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer?.
If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer?.
Solutions for If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
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