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  4. If x2-5 x-14>0 ⇒ x lie outside [α, β], then (α/β)=

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Q. If $x^2-5 x-14>0 \Rightarrow x$ lie outside $[\alpha, \beta]$, then $\frac{\alpha}{\beta}=$

TS EAMCET 2021

Solution:

$x^2-5 x-14>0 $
$ \Rightarrow(x+2)(x-7)>0$
$ x<-2 \text { or } x>7 \Rightarrow \alpha=-2 \text { or } \beta=7 $
$ \frac{\alpha}{\beta}=\frac{-2}{7}$