Let a, b and c be three positive real numbers | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the roots of the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ to be real,the discriminant $D$ must be greater than or equal to $0$.

Vedclass · Accepted Answer

$\lambda < \frac{4}{3}$

Vedclass · Answer

(C) For the roots of the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ to be real,the discriminant $D$ must be greater than or equal to $0$. $D = [2(a+b+c)]^2 - 4(1)(3\lambda(ab+bc+ca)) \geq 0$ $4(a+b+c)^2 - 12\lambda(ab+bc+ca) \geq 0$ $(a+b+c)^2 \geq 3\lambda(ab+bc+ca)$ $a^2+b^2+c^2+2(ab+bc+ca) \geq 3\lambda(ab+bc+ca)$ Dividing by $(ab+bc+ca)$,we get: $\frac{a^2+b^2+c^2}{ab+bc+ca} + 2 \geq 3\lambda$ Since $a, b, c$ are sides of a triangle,$a+b > c$,$b+c > a$,and $c+a > b$. It is a known inequality that $a^2+b^2+c^2 Thus,$\frac{a^2+b^2+c^2}{ab+bc+ca} Substituting this into the inequality: $3\lambda \leq \frac{a^2+b^2+c^2}{ab+bc+ca} + 2 $3\lambda < 4 \Rightarrow \lambda < \frac{4}{3}$.

Let $a, b$ and $c$ be three positive real numbers such that the sum of any two of them is greater than the third. All the values of $\lambda$ such that the roots of the equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ are real,are given by

Similar Questions

One root of the given equation $2x^5 - 14x^4 + 31x^3 - 64x^2 + 19x + 130 = 0$ is

The number of all $2$-digit numbers $n$ such that $n$ is equal to the sum of the square of the digit in its tens place and the cube of the digit in its units place is

If the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots $\alpha, \beta, \gamma, \delta$,then the values of $a$ and $b$ are:

The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$ is equal to:

The quadratic equation with real coefficients whose one root is $7 + 5i$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module