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

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(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

$\frac{-2}{7}$

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(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 $\frac{\alpha}{\beta}$.

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