Multiple choice

Let $a, b, c$ be the lengths of sides of a scalene triangle. If the roots of the equation $\displaystyle x^{2}+2(a+b+c)x+3\lambda (ab+bc+ca)=0$ $(\lambda \in R)$ are real then

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

For roots to be real, discriminant D >= 0. D = [2(a+b+c)]^2 - 4(1)(3*lambda*(ab+bc+ca)) >= 0. 4(a+b+c)^2 - 12*lambda*(ab+bc+ca) >= 0. (a^2+b^2+c^2+2ab+2bc+2ca) - 3*lambda*(ab+bc+ca) >= 0. Since a, b, c are sides of a triangle, we use the property that (a+b+c)^2 > 3(ab+bc+ca). The condition leads to lambda < 4/3.

AI explanation

For real roots, the discriminant must be non-negative, giving [2(a+b+c)]^2 - 12(lambda)(ab+bc+ca) >= 0. This simplifies to (a+b+c)^2 >= 3(lambda)(ab+bc+ca). Using the identity (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca), we get a^2+b^2+c^2 >= (3(lambda)-2)(ab+bc+ca). Since a, b, and c are sides of a scalene triangle, the triangle inequality ensures (a^2+b^2+c^2)/(ab+bc+ca) is always strictly less than 4/3. Therefore, (3(lambda)-2) must be less than 4/3, which restricts lambda to values smaller than 4/3. The answer is lambda < 4/3.

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