Multiple choice

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

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

For roots to be real and distinct, D > 0. D = [2(a+b+c)]^2 - 4(1)(3*lambda*(ab+bc+ca)) > 0. 4(a^2+b^2+c^2+2ab+2bc+2ca) - 12*lambda*(ab+bc+ca) > 0. (a^2+b^2+c^2) + 2(ab+bc+ca) - 3*lambda*(ab+bc+ca) > 0. Since (a-b)^2+(b-c)^2+(c-a)^2 > 0, we use the identity a^2+b^2+c^2 > ab+bc+ca. The condition simplifies to lambda < 4/3.

AI explanation

For roots to be real and distinct, the discriminant must be positive, so 4 times the square of (a plus b plus c) minus 12 lambda times (ab plus bc plus ca) is greater than 0. Expanding the squared term gives a squared plus b squared plus c squared plus 2ab plus 2bc plus 2ca, which allows simplifying the inequality to (a squared plus b squared plus c squared) minus (ab plus bc plus ca) is greater than 3 lambda times (ab plus bc plus ca). Because a, b, and c are sides of a triangle, a squared plus b squared plus c squared is strictly less than 2 times (ab plus bc plus ca). Substituting this condition gives (a squared plus b squared plus c squared) divided by (ab plus bc plus ca) is less than 4 thirds, which implies lambda is less than 4 thirds.

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