Multiple choice

Let $\mathrm{a},\ \mathrm{b},\ \mathrm{c}$ be the sides of a triangle . No two of them are equal and $\lambda\in R$ . If the roots of the equation $\mathrm{x}^{2}+2(\mathrm{a}+\mathrm{b}+\mathrm{c})\mathrm{x} +3\lambda (\mathrm{a}\mathrm{b}+ \mathrm{b}\mathrm{c}+\mathrm{c}\mathrm{a})=0$ are real, then

  1. $\displaystyle \lambda<\frac{4}{3}$
  2. $\displaystyle \lambda>\frac{5}{3}$
  3. $\displaystyle \lambda\in(\frac{1}{3}, \displaystyle \frac{5}{3})$
  4. $\displaystyle \lambda\in(\frac{4}{3},\frac{5}{3})$
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A Correct answer
Explanation

For the quadratic equation to have real roots, the discriminant D = 4(a+b+c)^2 - 12*lambda*(ab+bc+ca) must be >= 0. This simplifies to (a+b+c)^2 >= 3*lambda*(ab+bc+ca). Using the inequality (a+b+c)^2 > 3(ab+bc+ca) for distinct a, b, c, we find lambda < (a+b+c)^2 / (3(ab+bc+ca)). Since (a+b+c)^2 / (ab+bc+ca) > 3, the condition holds for lambda < 4/3.

AI explanation

For the roots to be real, the discriminant must be non-negative, so we evaluate [2(a+b+c)]^2 - 4(1)[3*lambda*(ab+bc+ca)] >= 0. This simplifies to 4(a^2+b^2+c^2+2ab+2bc+2ca) - 12*lambda*(ab+bc+ca) >= 0. Factoring this yields 4[(a-b)^2+(b-c)^2+(c-a)^2] >= 12*lambda*(ab+bc+ca). Because a, b, and c are distinct sides of a triangle, the term [(a-b)^2+(b-c)^2+(c-a)^2] is strictly positive, leading to lambda < 4/3.

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