Complex Numbers And Quadratic Equations question 759
Question: Both the roots of the given equation $ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 $ are always [MNR 1986; IIT 1980; Kurukshetra CEE 1998; RPET 2002]
Options:
A) Positive
B) Negative
C) Real
D) Imaginary
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation $ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 $ can be re-written as $ 3x^{2}-2(a+b+c)x+(ab+bc+ca)=0 $ $ \Delta =4{ {{(a+b+c)}^{2}}-3(ab+bc+ca) }(\because b^{2}-4ac=\Delta ) $ $ =4(a^{2}+b^{2}+c^{2}-ab-bc-ac) $ $ =2{ {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}} }\ge 0 $ Hence both roots are always real.
BETA
