Question
Both the roots of the equation $$\left( {x - b} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0$$ are always
A.
positive
B.
real
C.
negative
D.
none of these.
Answer :
real
Solution :
The given equation is
$$\eqalign{
& \,\,\,\,\,\,\left( {x - b} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0 \cr
& \Rightarrow \,\,3{x^2} - 2\left( {a + b + c} \right)x + \left( {ab + bc + ca} \right) = 0 \cr
& \,\,\,\,\,\,{\text{Discriminant}} = 4{\left( {a + b + c} \right)^2} - 12\left( {ab + bc + ca} \right) \cr
& \,\,\,\,\,\,\, = 4\left[ {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right] \cr
& \,\,\,\,\,\,\, = 2\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \geqslant 0\,\,\forall \,\,a,b,c \cr
& \therefore {\text{ Roots of given equation are always real}}{\text{.}} \cr} $$