Theory of Equations 1 Question 35

36. Both the roots of the equation

(x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=0

are always

(1980, 1M)

(a) positive

(b) negative

(c) real

(d) None of the above

Show Answer

Answer:

Correct Answer: 36. (c)

Solution:

  1. (x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0

⇒3x2−2(a+b+c)x+(ab+bc+ca)=0

Now, discriminant =4(a+b+c)2−12(ab+bc+ca)

=4a2+b2+c2−ab−bc−ca=2(a−b)2+(b−c)2+(c−a)2

which is always positive.

Hence, both roots are real.

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