Question

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

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

Answer 1

The correct option is A Real

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

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

Now, here $A=3,B=-2(a+b+c)$

$C=ab+bc+ca$

Therefore, $D=\sqrt{B^2-4AC}$

$=\sqrt{(-2(a+b+c){)}^2-4(3\left)\right(ab+bc+ca)}$

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

$=2\sqrt{a^2+b^2+c^2-ab-bc-ca}$

$=2\sqrt{\frac{1}{2}\left\{\right(a-b{)}^2-(b-c{)}^2+(c-a{)}^2\}}\ge 0$

This is always $\ge 0$, we have real roots for the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$