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The least integral value $\alpha$ of $x$ such that $\frac{x-5}{x^2+5 x-14}>0$, satisfies :
Options:
Solution:
2116 Upvotes
Verified Answer
The correct answer is:
$\alpha^2+3 \alpha-4=0$
$\alpha^2+3 \alpha-4=0$
$$
\text { } \begin{aligned}
& \frac{x-5}{x^2+5 x-14}>0 \Rightarrow x^2+5 x-14 < x-5 \\
& \Rightarrow x^2+4 x-9 < 0 \\
& \Rightarrow \alpha=-5,-4,-3,-2,-1,0,1
\end{aligned}
$$
$\alpha=-5$ does not satisfy any of the options
$\alpha=-4$ satisfy the option (a) $\alpha^2+3 \alpha-4=0$
\text { } \begin{aligned}
& \frac{x-5}{x^2+5 x-14}>0 \Rightarrow x^2+5 x-14 < x-5 \\
& \Rightarrow x^2+4 x-9 < 0 \\
& \Rightarrow \alpha=-5,-4,-3,-2,-1,0,1
\end{aligned}
$$
$\alpha=-5$ does not satisfy any of the options
$\alpha=-4$ satisfy the option (a) $\alpha^2+3 \alpha-4=0$
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