Aspire Faculty ID #10472 · Topic: NIMCET 2014 · Just now
NIMCET 2014

If $x^{2} + 2ax + 10 - 3a > 0$ for all x ∈ R, then

Solution

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: $a \in (-5, 2)$
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