Question

Both the roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always

• A. positive
• B. negative
• C. real
• D. None of these

Answer 1

Answer: (C)

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
$\Rightarrow 3x^2-2(a+b+c)x+(ab+bc+ca)=0$
Now, discriminant $=4(a+b+c{)}^2-12(ab+bc+ca)$
$=4\{a^2+b^2+c^2-ab-bc-ca\}$
$=2\{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}$
which is always positive.
Hence, both roots are real.