Find all values of $a$ for which the inequality $(a + 4) x^{2} - 2 a x + 2 a - 6 < 0$ is satisfied for all $x ϵ . R .$

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Solution

$(a + 4) x^{2} - 2 a x + 2 a - 6 < 0 \forall x ϵ R$
$a + 4 < 0$
$a < - 4, a ϵ (- \infty, - 4)$
$b < 0$
$(x - a)^{2} - 4 (a + 4) (2 a - 6) < 0$
$4 a^{2} - 8 (a + 4) (a - 3) < 0$
$a^{2} - 2 (a^{2} + a - 12) < 0$
$a^{2} - 2 a^{2} - 2 a + 24 < 0$
$a^{2} + 2 a - 24 > 0$
$(a + 6) (a - 4) > 0$
$a ϵ (- \infty, - 6) \cup (4, \infty)$
So, $a ϵ (- \infty, - 6)$

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