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If $x^2+2 p x-2 p+8>0$ for all real values of $x$, then the set of all possible values of $p$ is
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Verified Answer
The correct answer is:
$(-4,2)$
Given, $x^2+2 p x-2 p+8>0$, for all real values of $x$.
$$
\begin{aligned}
\therefore \quad & D=b^2-4 a c < 0 \\
& \quad[\because \text { if a }>0 \text { and } D < 0, \text { then } f(x)>0 \text { for all } x \in R] \\
\Rightarrow & (2 p)^2+4(2 p-8) < 0 \Rightarrow 4 p^2+8 p-32 < 0
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad p^2+2 p-8 < 0 \Rightarrow p^2+4 p-2 p-8 < 0 \\
& \Rightarrow \quad(p+4)(p-4 < 0
\end{aligned}
$$
$$
\Rightarrow \quad-4 < p < 2 \Rightarrow p \in(-4,2)
$$
$$
\begin{aligned}
\therefore \quad & D=b^2-4 a c < 0 \\
& \quad[\because \text { if a }>0 \text { and } D < 0, \text { then } f(x)>0 \text { for all } x \in R] \\
\Rightarrow & (2 p)^2+4(2 p-8) < 0 \Rightarrow 4 p^2+8 p-32 < 0
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad p^2+2 p-8 < 0 \Rightarrow p^2+4 p-2 p-8 < 0 \\
& \Rightarrow \quad(p+4)(p-4 < 0
\end{aligned}
$$
$$
\Rightarrow \quad-4 < p < 2 \Rightarrow p \in(-4,2)
$$
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