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 always

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

Answer 1

Answer: (C)

Given equation can be rewritten as

$3x^{2} - 2x\left(a + b + c\right)+ ab+ bc + ca = 0$

Now, Discriminant,

$D = 4\left(a + b + c\right)^{2} - 4\cdot3\left(ab + bc + ca\right)$

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

$=2\left[\left(a-b\right)^{2}+\left(b-c\right)^{2}+\left(c-a\right)^{2}\right] \ge\,0$

Hence, roots are real