Let's solve the given equations step-by-step.
### 1. The equation [tex]\(8 - 4x = 0\)[/tex]
First, let's isolate [tex]\(x\)[/tex]:
[tex]\[ 8 - 4x = 0 \][/tex]
Subtract 8 from both sides:
[tex]\[ -4x = -8 \][/tex]
Now, divide both sides by -4:
[tex]\[ x = \frac{-8}{-4} \][/tex]
[tex]\[ x = 2 \][/tex]
So, the equation [tex]\(8 - 4x = 0\)[/tex] has one real solution, which is [tex]\(x = 2\)[/tex].
### 2. The equation [tex]\(2 - 9 = 0\)[/tex]
We can simplify the left-hand side:
[tex]\[ 2 - 9 = -7 \][/tex]
This simplifies to:
[tex]\[ -7 = 0 \][/tex]
This equation is clearly false because [tex]\(-7\)[/tex] is never equal to [tex]\(0\)[/tex]. Therefore, there are no real solutions for the equation [tex]\(2 - 9 = 0\)[/tex].
### 3. The equation [tex]\(x^3 + 3x^2 + 5x + 15 = 0\)[/tex]
To solve this cubic equation, we will explore its roots.
A cubic equation [tex]\(ax^3 + bx^2 + cx + d = 0\)[/tex] may have up to three roots, which can be real or complex (imaginary).
For the equation [tex]\(x^3 + 3x^2 + 5x + 15 = 0\)[/tex]:
- There is one real solution to this equation.
- There are two imaginary solutions.
Thus, the equation [tex]\(x^3 + 3x^2 + 5x + 15 = 0\)[/tex] has one real solution and two imaginary solutions.
### Summary:
1. The equation [tex]\(8 - 4x = 0\)[/tex] has one real solution, which is [tex]\(x = 2\)[/tex].
2. The equation [tex]\(2 - 9 = 0\)[/tex] has no real solutions.
3. The equation [tex]\(x^3 + 3x^2 + 5x + 15 = 0\)[/tex] has one real solution and two imaginary solutions.
