if the domain of function below is $$ D f=\mathbb{R}-\{-2\} $$, what is $$ a+b= $$ ? $$ f(x)=\frac{x^{2}-1}{x^{2}+a x+b} $$

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To determine the values of $$ a $$ and $$ b $$ such that the function

$$
f(x) = \frac{x^2 - 1}{x^2 + ax + b}
$$

has a domain of $$ D_f = \mathbb{R} - \{-2\} $$, we need to ensure that the denominator $$ x^2 + ax + b $$ is equal to zero at $$ x = -2 $$ and is non-zero for all other real numbers.

First, we substitute $$ x = -2 $$ into the denominator:

$$
(-2)^2 + a(-2) + b = 0
$$

This simplifies to:

$$
4 - 2a + b = 0
$$

Rearranging gives us:

$$
b = 2a - 4 \quad \text{(1)}
$$

Next, we need to ensure that the quadratic $$ x^2 + ax + b $$ does not have any other real roots. The condition for a quadratic $$ Ax^2 + Bx + C $$ to have no real roots is that its discriminant must be less than zero. The discriminant $$ D $$ for our quadratic is given by:

$$
D = a^2 - 4b
$$

Substituting $$ b $$ from equation (1) into the discriminant:

$$
D = a^2 - 4(2a - 4)
$$

This simplifies to:

$$
D = a^2 - 8a + 16
$$

We want this discriminant to be less than zero:

$$
a^2 - 8a + 16 < 0
$$

Factoring the quadratic:

$$
(a - 4)^2 < 0
$$

The expression $$ (a - 4)^2 $$ is always non-negative and equals zero when $$ a = 4 $$. Therefore, the only solution is:

$$
a = 4
$$

Substituting $$ a = 4 $$ back into equation (1) to find $$ b $$:

$$
b = 2(4) - 4 = 8 - 4 = 4
$$

Thus, we have:

$$
a = 4 \quad \text{and} \quad b = 4
$$

Finally, we calculate $$ a + b $$:

$$
a + b = 4 + 4 = 8
$$

Therefore, the answer is:

$$
\boxed{8}
$$
$$ a+b=8 $$

if the domain of function below is
, what is ?

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