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Mathematics Question on Complex Numbers and Quadratic Equations

Both the roots of the equation (x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=0(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

real

Explanation

Solution

(x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=0(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0
⇒3x2−2(a+b+c)x+(ab+bc+ca)=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+b+c)^2-12(ab+bc+ca)
=4(a2+b2+c2−ab−bc−ca)=4\big(a^2+b^2+c^2-ab-bc-ca\big)
=2\Big\\{(a-b)^2+(b-c)^2+(c-a)^2\Big\\}
which is always positive.
Hence, both roots are real.