Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.1k views
by (645 points)
Roots of the equation

a(b-c)x² + b(c-a)x + c(a-b) = 0

are equal and real.

Prove that,

1/a + 1/c  =  2/b

1 Answer

+2 votes
by (10.9k points)
edited by
 
Best answer
As the roots of the given quadratic equation are equal and real then its discriminant will be zero

[b(c-a)]^2=4×a(b-c)×c(a-b)

=>(bc)^2+(ab)^2+2ab^2c=4[(ab-ca))(ca-bc)]

=>(bc)^2+(ab)^2-2ab^2c=4a^2bc-4ab^2c +4abc^2-4(ca)^2

=>(bc)^2+(ab)^2+(2ca)^2+2ab^2c-4abc^2-4a^2bc=0

=>(bc+ab-2ca)^2=0

=>bc+ab=2ca

=>1/a+1/c=2/b ( dividing both sides by abc)

Proved

Related questions

+14 votes
1 answer
+3 votes
0 answers
0 votes
1 answer
asked Aug 24, 2023 in Mathematics by Yaksh. (95 points)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...