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WBJEE · Maths · Quadratic Equation

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

  1. A positive
  2. B negative
  3. C real
  4. D imaginary
Verified Solution

Answer & Solution

Correct Answer

(C) real

Step-by-step Solution

Detailed explanation

Hints : \(3 x^2-2 x(a+b+c)+a b+b c+c a=0\) \[ \begin{aligned} & \mathrm{D}=4(a+b+c)^2-4.3(a b+b c+c a) \\ & =4\left(a^2+b^2+c^2-a b-b c-c a\right) \\ & =2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\ & =\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\ & \geq 0 \end{aligned} \]