Question

For $a,b,c\in R$ both the roots of the equation $(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$ are

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

Answer 1

The correct option is C real

$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$
$3x^2-2(a+b+c)x+ab+bc+ca=0$
Consider $D=b^2-4ac$
$=4(a+b+c{)}^2-4(3\left)\right(ab+bc+ca)$
$=4(a^2+b^2+c^2-ab-bc-ca)$
$=4\left(\frac{1}{2}\right[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2])$
$=2\left(\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2)\ge 0$
$\therefore$ Roots are real .