🎥 Video solution / Text Solution of this question is given below:
Given quadratic inequality:
$x^2 + 2ax + (10 - 3a) > 0 \quad \text{for all } x \in \mathbb{R}$
Condition for a Quadratic to be Strictly Positive:
For any quadratic expression
$Ax^2 + Bx + C > 0$ for all $x \in \mathbb{R}$:
1. $A > 0$ (parabola opens upwards)
2. $D = B^2 - 4AC < 0$ (the parabola does not touch or intersect the x-axis)
Here:
$A = 1 > 0$ (satisfied)
$B = 2a$
$C = 10 - 3a$
Applying the Discriminant Condition ($D < 0$):
$D = (2a)^2 - 4(1)(10 - 3a) < 0$
$4a^2 - 4(10 - 3a) < 0$
Divide both sides by $4$:
$a^2 - (10 - 3a) < 0$
$a^2 + 3a - 10 < 0$
Factor the quadratic equation in $a$:
$(a + 5)(a - 2) < 0$
Using the wavy curve method / sign scheme:
$-5 < a < 2$
or in interval notation:
$a \in (-5, 2)$
