Find the least integral positive value of $x$ which satisfied the inequality $\sqrt{x^{2} + 16 x + 64} > 20.$

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Solution

$x^{2} + 16 x + 64 \geq 0$
$(x + 8)^{2} \geq 0$
$x ϵ R$
$\sqrt{x^{2} + 16 x + 64} > 20$
Squarring both sides
$x^{2} + 16 x + 64 > 400$
$x^{2} + 16 x - 336 > 0$
$(x + 28) (x - 12) > 0$
$x ϵ (- \infty, - 28) \cup (12, \infty)$
So, least positive integral value $= 13$

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