Question

If $a,b,c\in R^{+}$, then both the roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always

• A. real and negative
• B. imaginary
• C. real and positive
• D. real and equal

Answer 1

The correct option is C real and positive

$(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$
$\Rightarrow 3x^2-2x(a+b+c)+ab+bc+ca=0$

Now, for the nature of roots
$D=4(a+b+c{)}^2-12(ab+bc+ca)=4a^2+4b^2+4c^2-4(ab+bc+ca)=2\left[\right(a-b{)}^2+(b-c{)}^2+(a-c{)}^2]\therefore D\ge 0$
$\Rightarrow$ Roots are real.

Let the roots be $\alpha ,\beta$
$\alpha +\beta =\frac{2}{3}(a+b+c)>0\cdots \left(1\right)\alpha \beta =\frac{1}{3}(ab+bc+ca)>0\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
we can conclude that roots are positive.