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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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The correct answer is:
$(-4,2)$
For any quadratic expression $\mathrm{P}(\mathrm{x})$
$\begin{aligned} & \mathrm{P}(\mathrm{x})>0 \forall \mathrm{x} \Rightarrow \mathrm{D} < 0, \mathrm{a}>0 \\ & \mathrm{D}=(2 \mathrm{p})^2-4(8-2 \mathrm{p}) < 0 \\ & 4 \mathrm{p}^2-32+8 \mathrm{p} < 0 \\ & (\mathrm{p}+4)(\mathrm{p}-2) < 0 \\ & \therefore \mathrm{p} \in(-4,2)\end{aligned}$
$\begin{aligned} & \mathrm{P}(\mathrm{x})>0 \forall \mathrm{x} \Rightarrow \mathrm{D} < 0, \mathrm{a}>0 \\ & \mathrm{D}=(2 \mathrm{p})^2-4(8-2 \mathrm{p}) < 0 \\ & 4 \mathrm{p}^2-32+8 \mathrm{p} < 0 \\ & (\mathrm{p}+4)(\mathrm{p}-2) < 0 \\ & \therefore \mathrm{p} \in(-4,2)\end{aligned}$
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