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.
Let $\alpha$ be a solution of $x^2 + x + 1 = 0$, and for some $a$ and $b$ in $\mathbb{R}$, $ \begin{bmatrix} 1 & 16 & 13 \\-1 & -1 & 2 \\-2 & -14 & -8 \end{bmatrix} \begin{bmatrix} 4 \\a \\b \end{bmatrix} = \begin{bmatrix} 0 \\0 \\0 \end{bmatrix}. $ If $\frac{4}{\alpha^4} + \frac{m} {\alpha^a} + \frac{n}{\alpha^b} = 3$, then $m + n$ is equal to _____.
If the set of all values of \( a \), for which the equation \( 5x^3 - 15x - a = 0 \) has three distinct real roots, is the interval \( (\alpha, \beta) \), then \( \beta - 2\alpha \) is equal to