Question

Both the roots of the given 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

Given equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ can be written as
$3x^2-2(a+b+c)x+(ab+bc+ca)=0$

Now, $D=B^2-4AC=4\left\{\right(a+b+c{)}^2-3(ab+bc+ca)\}$

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

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

$=2\left\{\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}\ge 0$ as sum of squares is always greater than or equal to zero

$\Rightarrow D\ge 0$

Hence, both roots are always real .