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Mathematics Question on Complex Numbers and Quadratic Equations

If a, b,c are the sides of a triangle ABC such that x2−2(a+b+c)x+3λ(ab+bc+ca)=0x^2 - 2(a+b+c)x+3\lambda(ab+bc+ca)=0 has real roots, then

A

$\lambda

B

λ>53\lambda>\frac{5}{3}

C

λ∈(43,53)\lambda\in\big(\frac{4}{3},\frac{5}{3}\big)

D

λ∈(13,53)\lambda\in\big(\frac{1}{3},\frac{5}{3}\big)

Answer

$\lambda

Explanation

Solution

Since, roots are real, therefore D≥0D\ge0
⇒   4(a+b+c)2−12λ(ab+bc+ca)≥0\Rightarrow\, \, \, 4(a+b+c)^2-12\lambda(ab+bc+ca)\ge0
⇒         (a+b+c)2≥3λ(ab+bc+ca)\Rightarrow\, \, \, \, \, \, \, \, \, (a+b+c)^2\ge3\lambda(ab+bc+ca)
⇒         (a2+b2+c2)≥(ab+bc+ca)(3λ−2)\Rightarrow\, \, \, \, \, \, \, \, \, (a^2+b^2+c^2)\ge(ab+bc+ca)(3\lambda-2)
⇒         3λ−2≤a2+b2+c2ab+bc+ca        ...(i)\Rightarrow\, \, \, \, \, \, \, \, \, 3\lambda-2 \le \frac{a^2+b^2+c^2}{ab+bc+ca}\, \, \, \, \, \, \, \, ...(i)
Also, cosA=b2+c2−a22bc<1⇒b2+c2−a2<2bccosA=\frac{b^2+c^2-a^2}{2bc}<1 \Rightarrow b^2+c^2-a^2<2bc
Similarly, \hspace15mm\, \, \, c^2+a^2-b^2<2ca
and \hspace25mm\, \, \, a^2+b^2-c^2<2ab
\Rightarrow\hspace25mm a^2+b^2+c^2<2(ab+bc+ca)
\Rightarrow\hspace25mm\frac{a^2+b^2+c^2}{ab+bc+ca}<2\, \, \, \, \, \, \, \, \, \, \, ...(ii)
From Eqs. (i) and (ii), we get
$\hspace25mm 3\lambda-2<2 \Rightarrow \lambda