First slide

Theory of expressions

Question

If $x^{2} + 2 a x + 10 - 3 a > 0$ for each $x \in R$
then

Moderate

A

$a < - 5$
B

$- 5 < a < 2$
C

$a > 5$
D

$2 < a < 5$

Solution

$x^{2} + 2 a x + 10 - 3 a > 0 \forall x \in R$
$\begin{array}{ll} \Rightarrow (x + a)^{2} - (a^{2} + 10 - 3 a) > 0 \forall x \in R \\ \Rightarrow a^{2} + 3 a - 10 < 0 \\ \Rightarrow (a + 5) (a - 2) < 0 \\ \Rightarrow - 5 < a < 2 \end{array}$

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