Question

Find the set of values of a for which $9x^{2}+2(a+1)x+4>0$ $\forall x \in R$

Answer 1

$a \in (-7, 5)$

Answer 2

Given the quadratic

$$f(x)=9x^2+2(a+1)x+4,$$

for $f(x) > 0$ for all $x \in \mathbb{R}$, the necessary condition is that its discriminant must be negative.

1. Calculate the discriminant:

   $$D = [2(a+1)]^2 - 4(9)(4) = 4(a+1)^2 - 144.$$

2. Set the discriminant less than zero:

   $$4(a+1)^2 - 144 < 0.$$

3. Simplify:

   $$(a+1)^2 < 36.$$

4. Solve the inequality:

   $$-6 < a+1 < 6 \quad \Longrightarrow \quad -7 < a < 5.$$

Thus, the quadratic remains positive for all $x$ if and only if

$$\boxed{-7 < a < 5.}$$

Core Explanation:
For $9x^2+2(a+1)x+4 > 0$ to hold for every $x$, its discriminant must be negative. This results in the inequality $(a+1)^2 < 36$ which simplifies to $-7 < a < 5$.