Question

If $a,b,c$ are real, then 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. non-real

Answer 1

The correct option is C real

$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$
$\Rightarrow 3x^2-2(a+b+c)x+(bc+ca+ab)=0$
$\Delta =4(a+b+c{)}^2-12(bc+ca+ab)$
$=4\left[\right(a+b+c{)}^2-3(bc+ca+ab)]$
$=4[a^2+b^2+c^2-bc-ca-ab]$
$\Delta =2\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$
$\Rightarrow \Delta \ge 0$
$\Rightarrow$ Roots are real.