(A) Given the inequality: $x^2-5x-14 > 0$ Factorizing the quadratic expression: $(x+2)(x-7) > 0$ For the product to be positive,$x$ must be less than

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

(A) Given the inequality: $x^2-5x-14 > 0$ Factorizing the quadratic expression: $(x+2)(x-7) > 0$ For the product to be positive,$x$ must be less than the smaller root or greater than the larger root: $x \in (-\infty, -2) \cup (7, \infty)$ The problem states that $x$ lies outside the interval $[\alpha, \beta]$. Comparing the intervals,we identify $\alpha = -2$ and $\beta = 7$. Therefore,the value of $\frac{\alpha}{\beta} = \frac{-2}{7}$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic inequations and Newton Formula

Easy

If $x^2-5x-14 > 0$ implies $x$ lies outside $[\alpha, \beta]$,then find the value of $\frac{\alpha}{\beta}$.

TS EAMCET 2021 All TS EAMCET

A
$\frac{-2}{7}$
B
$\frac{-7}{2}$
C
$\frac{2}{7}$
D
$\frac{7}{2}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

4-2.Quadratic Equations and Inequations

Subtopic

Solution of quadratic inequations and Newton Formula

Previous Year

TS EAMCET 2021 Paper All TS EAMCET years →

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is non-negative for all positive real numbers $x$,then what must be the minimum value of $m$?

Difficult

View Solution

Let $S$ be the set of positive integral values of $a$ for which $\frac{ax^2+2(a+1)x+9a+4}{x^2-8x+32} < 0, \forall x \in R$. Then,the number of elements in $S$ is:

DifficultJEE MAIN 2024

View Solution

If $\alpha_1, \alpha_2$ and $\alpha_3$ are the roots of $x^3+3x+2=0$,then $\alpha_1^5+\alpha_2^5+\alpha_3^5=$

MediumAP EAMCET 2023

View Solution

The set $\{x \in R: 4+11x-3x^2>0\}$ is the interval

EasyAP EAMCET 2023

View Solution

If $x$ is a real number,for what values of $x$ does $3x^2 + 14x + 11 > 0$ hold?

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If $x^2-5x-14 > 0$ implies $x$ lies outside $[\alpha, \beta]$,then find the value of $\frac{\alpha}{\beta}$.

Similar Questions

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is non-negative for all positive real numbers $x$,then what must be the minimum value of $m$?

Let $S$ be the set of positive integral values of $a$ for which $\frac{ax^2+2(a+1)x+9a+4}{x^2-8x+32} < 0, \forall x \in R$. Then,the number of elements in $S$ is:

If $\alpha_1, \alpha_2$ and $\alpha_3$ are the roots of $x^3+3x+2=0$,then $\alpha_1^5+\alpha_2^5+\alpha_3^5=$

The set $\{x \in R: 4+11x-3x^2>0\}$ is the interval

If $x$ is a real number,for what values of $x$ does $3x^2 + 14x + 11 > 0$ hold?

Vedclass Test Series

Exam Paper Generator

Online Exam Module