If x^2-5x-14 0 implies x lies outside [alph | vedclass

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

Question

(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

Vedclass · Accepted Answer

$\frac{-2}{7}$

Vedclass · Answer

(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 $\frac{\alpha}{\beta} = \frac{-2}{7}$.

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

Similar Questions

The set of solutions satisfying both $x^2+5x+6 \geq 0$ and $x^2+3x-4 < 0$ is

For the inequality $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$,the set of real values of $x$ is:

If $x^2+2px-2p+8>0$ for all real values of $x$,then the set of all possible values of $p$ is

Let $f(x) = x^2 + 4x + 1$. Then

If $[x]$ denotes the greatest integer not exceeding $x$,then the values of $x$ satisfying $[x]^2-7[x]+12 \leq 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module