How many integers satisfy the inequality
\left| \frac { 10(x+1) } {x^2 + 2x + 3 } \right| \geq 1?
10(x+1)/(x^2+2x+3) >= 1
since x^2+2x+3 > 0,
10x+10 >= x^2+2x+3
x^2 - 8x - 7 <= 0
So, all integers n such that
4-√23 <= n <= 4+√23
To solve the inequality
\left| \frac { 10(x+1) } {x^2 + 2x + 3 } \right| \geq 1,
we can start by multiplying both sides by the denominator, x^2 + 2x + 3, while preserving the sign:
\left| 10(x+1) \right| \geq x^2 + 2x + 3.
Next, we can split the inequality into two separate cases depending on the sign of x^2 + 2x + 3:
Case 1: x^2 + 2x + 3 > 0
In this case, we can remove the absolute value signs without changing the inequality:
10(x+1) \geq x^2 + 2x + 3.
Expanding and simplifying gives us a quadratic equation:
x^2 - 8x + 7 \leq 0.
Factoring the quadratic equation, we get:
(x - 1)(x - 7) \leq 0.
To solve this inequality, we use the sign chart method and test different intervals, checking the sign of the expression (x - 1)(x - 7):
Interval (-∞, 1]:
(x - 1)(x - 7) < 0
Solving this inequality, we find that x lies in the interval (1, 7].
So for Case 1, the values of x that satisfy the inequality are x \in (1, 7].
Case 2: x^2 + 2x + 3 < 0
In this case, we need to flip the inequality when removing the absolute value signs:
10(x+1) \leq -(x^2 + 2x + 3).
Expanding and simplifying gives us the quadratic equation:
x^2 + 12x + 7 \geq 0.
Using the same method as before, we find that x lies in the interval (-∞, -3] \cup [-4, ∞).
So for Case 2, the values of x that satisfy the inequality are x \in (-∞, -3] \cup [-4, ∞).
Combining the solutions from both cases, we find that the integers satisfying the inequality are x \in \{1, 2, 3, 4, 5, 6, 7\}.