We are given the equation:
\[
|x + 1| |x + 3| - 4|x + 2| + 5 = 0
\]
Case 1: \(x \leq -3\)
Replacing all absolute values with negatives:
\[
(x + 1)(x + 3) + 4(x + 2) + 5 = 0
\]
Simplify:
\[
x^2 + 4x + 3 + 4x + 8 + 5 = 0
\]
\[
x^2 + 8x + 16 = 0
\]
\[
(x + 4)^2 = 0 \Rightarrow x = -4
\]
Case 2: \(-3 \leq x \leq -2\)
\[
-(x + 1)(x + 3) + 4(x + 2) + 5 = 0
\]
Simplify:
\[
-x^2 - 4x - 3 + 4x + 8 + 5 = 0
\]
\[
-x^2 + 10 = 0
\]
\[
x = \pm \sqrt{10}
\]
Since \(x \in [-3, -2]\), no valid root in this interval.
Case 3: \(-2 \leq x \leq -1\)
\[
-(x + 1)(x + 3) - 4(x + 2) + 5 = 0
\]
Simplify:
\[
-x^2 - 4x - 3 - 4x - 8 + 5 = 0
\]
\[
-x^2 - 8x - 6 = 0
\]
\[
x^2 + 8x + 6 = 0
\]
\[
x = \frac{-8 \pm \sqrt{64 - 24}}{2} = -4 \pm \sqrt{10}
\]
Case 4: \(x \geq -1\)
\[
(x + 1)(x + 3) - 4(x + 2) + 5 = 0
\]
Simplify:
\[
x^2 + 4x + 3 - 4x - 8 + 5 = 0
\]
\[
x^2 = 0 \Rightarrow x = 0
\]
Final Step:
Checking all valid roots, the solutions are \(x = -4\) and \(x = 0.\)
Hence, the number of solutions = 2.
Given equation:
\[|x + 1||x + 3| - 4|x + 2| + 5 = 0.\]
To solve this, we break it into different cases based on the values of \(x\) that change the absolute values:
Case 1: \(x \leq -3\)
\[(x + 1)(x + 3) + 4(x + 2) + 5 = 0.\]
Simplifying:
\[x^2 + 4x + 3 + 4x + 8 + 5 = 0 \implies x^2 + 8x + 16 = 0 \implies (x + 4)^2 = 0.\]
\[x = -4.\]
Case 2: \(-3<x \leq -2\)
\[-(x + 1)(x + 3) + 4(x + 2) + 5 = 0.\]
Simplifying:
\[-x^2 - 4x - 3 + 4x + 8 + 5 = 0 \implies -x^2 + 10 = 0 \implies x^2 = 10.\]
\[x = \pm \sqrt{10}.\]
Case 3: \(-2<x \leq -1\)
\[-(x + 1)(x + 3) - 4(x + 2) + 5 = 0.\]
Simplifying:
\[-x^2 - 4x - 3 - 4x - 8 + 5 = 0 \implies -x^2 - 8x - 6 = 0 \implies x^2 + 8x + 6 = 0.\]
Solving using the quadratic formula:
\[x = \frac{-8 \pm \sqrt{64 - 24}}{2} = \frac{-8 \pm \sqrt{40}}{2} = -4 \pm \sqrt{10}.\]
Case 4: \(x>-1\)
\[x^2 + 4x + 3 - 4x - 8 + 5 = 0.\]
Simplifying:
\[x^2 = 0 \implies x = 0.\]
The number of distinct real roots is:
Total Solutions = 2.
Answer: 2.
Designate whether each of the following compounds is aromatic or not aromatic.
