Question

If a, b and c are real, then both the roots of the equation $(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$ are always

• A. Positive
• B. Negative
• C. Real
• D. Imaginary

Answer 1

The correct option is C Real

Given equation is $(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$

We can further simplify this equation by multiplying terms and taking common powers of $x$.

$(x^2-bx-cx+bc)+(x^2-cx-ax+ac)+(x^2-ax-bx+ab)=0$

$\Rightarrow 3x^2-2ax-2bx-2cx+ab+bc+ca$
$\Rightarrow 3x^2-2x(a+b+c)+ab+bc+ca=0$
We know that, discriminant of a quadratic equation of the form $a{x}^2+bx+c=0$ is given by:

$D=b^2-4ac$

$\therefore D=[-2(a+b+c){]}^2-4\times 3\times (ab+bc+ca)$

$\Rightarrow D=4{(a+b+c)}^2-4\times 3(ab+bc+ca)$
$\Rightarrow 4(a^2+b^2+c^2-ab-bc-ca)$
$\Rightarrow 2[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$
Given that, $a,b\text{and}c$ are real.

Hence, $D\ge 0$
Therefore, roots are real.

Hence, option C is correct.