Algebraic Triangle.

Algebra Level 4

If $a,b,c$ are the sides of a triangle such that $x^2-2(a+b+c)x+3\lambda(ab+bc+ca)=0$ has real roots, which of the following is correct?

4/3<\lambda<5/3

NONE OF THESE

\lambda>5/3

\lambda<4/3

1 solution

Rudresh Tomar
Nov 13, 2014

$if\quad a,b,c\quad are\quad the\quad sides\quad of\quad a\quad triangle\quad then,\\ { (a+b+c) }^{ 2 }\ge 4(ab+bc+ca)\\ \Longrightarrow \frac { { (a+b+c) }^{ 2 } }{ (ab+bc+ca) } \ge 4\quad \dashrightarrow (1)\\ \\ now\quad \rightarrow \quad { x }^{ 2 }-2(a+b+c)x+3\lambda (ab+bc+ca)=0\\ has\quad real\quad roots\quad \therefore \\ D\ge 0\\ \Longrightarrow \quad 4{ (a+b+c) }^{ 2 }-4.3\lambda (ab+bc+ca)\ge 0\\ \Longrightarrow { \quad (a+b+c) }^{ 2 }-3\lambda (ab+bc+ca)\ge 0\\ \Longrightarrow \quad \frac { { (a+b+c) }^{ 2 } }{ (ab+bc+ca) } \ge 3\lambda \\ \Longrightarrow \quad 4\ge 3\lambda \\ \Longrightarrow \quad \lambda <\frac { 4 }{ 3 } \\$

It is a question of ''The pearson guide for complete mathematics for IIT-JEE"

Parth Lohomi - 6 years, 6 months ago

@Rudresh Tomar I don't know how your proof works, but there's an awful lot of mistakes in it. For example, the identity you assumed at the beginning does not hold true for a triangle with sides $a=3,b=4,c=5$ since $(a+b+c)^2=144$ and $4(ab+bc+ca)=4(12+20+15)=188$ and we can clearly see that $144\ngeq 188$ . Furthermore, the substitution you made at the before last line doesn't work when when you are dealing with inequalities. For example, we have $5\geq 1 \land 5\geq 3$ but $1\ngeq 3$ .

Prasun Biswas - 6 years, 6 months ago

not true bro :D, $4(ab+bc+ac) > (a+b+c)^2$ because $2(ab+bc+ac) \ge a^2+b^2+c^2$ beacause $ a(b+c-a)+b(a+c-b)+c(a+b-c) >0$

Road Human - 6 years, 5 months ago

Algebraic Triangle.

1 solution

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