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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Quadratic Equation

BITSAT 2016 · Mathematics · Quadratic Equation

If a, b and c are real numbers then the roots of the equation...

If a,ba, b and cc are real numbers then the roots of the equation (x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always

Options

  1. Areal
  2. Bimaginary
  3. Cpositive
  4. Dnegative

Correct answer

A. real

Step-by-step solution

Given equation is (x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0⇒3x2−2(b+a+c)x+ab+bc+ca=0(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0\Rightarrow 3 x^{2}-2(b+a+c) x+a b+b c+c a=0 Now, here A=3,B=−2(a+b+c)C=ab+bc+ca∴D=B2−4AC=(−2(a+b+c))2−4(3)(ab+bc+ca)=4(a+b+c)2−12(ab+bc+ca)=2a2+b2+c2−ab−bc−ca=212{(a−b)2−(b−c)2+(c−a)2}≥0A=3, B=-2(a+b+c)C=a b+b c+c a\therefore D=\sqrt{B^{2}-4 A C}=\sqrt{(-2(a+b+c))^{2}-4(3)(a b+b c+c a)}=\sqrt{4(a+b+c)^{2}-12(a b+b c+c a)}=2 \sqrt{a^{2}+b^{2}+c^{2}-a b-b c-c a}=2 \sqrt{\frac{1}{2}\left\{(a-b)^{2}-(b-c)^{2}+(c-a)^{2}\right\}} \geq 0
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