Question

Both the roots of the equation (x - a) ( x - b) + (x - b) (x - c) + (x - c) (x - a) = 0 are always

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

Answer 1

The correct option is C

Real

Explanation for correct option:

Step 1: Solving the equation:

Here, $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$

$\Rightarrow x^2-bx-ax+ab+x^2-bx-cx+bc+x^2-ax-cx+ac=0$

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

Step 2: Declaring the theorem:

We know, $D=B^2-4AC$

Step 3: Comparing the equations:

Here, $B=a+b+c$,

$A=3$

$C=ab+bc+ca$

$=4{(a+b+c)}^2-12(ab+bc+ca)$

$=4(a^2+b^2+c^2–ab–bc–ca)$

$=2(a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca)$

$=2{(a-b)}^2+{(b-c)}^2+{(c-a)}^2$

Since the sum of squares is always greater than or equal to zero,

$\Rightarrow D\ge 0$

Therefore, both roots will always be real.

Hence, Option (C) is correct.