Theory of Equations 1 Question 35
36. Both the roots of the equation
$(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$
are always
(1980, 1M)
(a) positive
(b) negative
(c) real
(d) None of the above
Show Answer
Solution:
- $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
$\Rightarrow \quad 3 x^{2}-2(a+b+c) x+(a b+b c+c a)=0$
Now, discriminant $=4(a+b+c)^{2}-12(a b+b c+c a)$
$$ \begin{aligned} & =4{a^{2}+b^{2}+c^{2}-a b-b c-c a } \\ & =2{(a-b)^{2}+(b-c)^{2}+(c-a)^{2} } \end{aligned} $$
which is always positive.
Hence, both roots are real.
