Consider the following numbers: $$ \sqrt{27} ; \sqrt[3]{-27} ; \sqrt{-27} $$. Which ONE of these numbers is: $$ \begin{array}{l}1.3 .1 \quad ext { Irrational } \ 1.3 .2 \quad ext { Non-real } \ ext { ESTION } 2 \ ext { Solve for } x ext { : } \ 2.1 .1 \quad 15 x^{2}-8=14 x \ 2.1 .2 \quad 5^{x}=\frac{1}{125}\end{array} $$

Question

Consider the following numbers: $$ \sqrt{27} ; \sqrt[3]{-27} ; \sqrt{-27} $$. Which ONE of these numbers is: $$ \begin{array}{l}1.3 .1 \quad 	ext { Irrational } \ 1.3 .2 \quad 	ext { Non-real } \ 	ext { ESTION } 2 \ 	ext { Solve for } x 	ext { : } \ 2.1 .1 \quad 15 x^{2}-8=14 x \ 2.1 .2 \quad 5^{x}=\frac{1}{125}\end{array} $$

UpStudy AI · Accepted Answer

Let's analyze the given numbers and solve the equations step by step.

### Part 1: Analyzing the Numbers

1. **Number 1: $$ \sqrt{27} $$**
   - This can be simplified as $$ \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3} $$.
   - Since $$ \sqrt{3} $$ is an irrational number, $$ \sqrt{27} $$ is also irrational.

2. **Number 2: $$ \sqrt[3]{-27} $$**
   - The cube root of $$-27$$ is $$-3$$ because $$(-3)^3 = -27$$.
   - This is a rational number.

3. **Number 3: $$ \sqrt{-27} $$**
   - This can be expressed as $$ \sqrt{-27} = \sqrt{27}i = 3\sqrt{3}i $$.
   - This is a non-real number because it involves the imaginary unit $$i$$.

### Summary of Part 1:
- **Irrational**: $$ \sqrt{27} $$
- **Non-real**: $$ \sqrt{-27} $$

### Part 2: Solving the Equations

#### Equation 2.1.1: $$ 15x^2 - 8 = 14x $$

Rearranging the equation:
$$
15x^2 - 14x - 8 = 0
$$

Now, we will use the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$ a = 15, b = -14, c = -8 $$.

Calculating the discriminant:
$$
b^2 - 4ac = (-14)^2 - 4 \cdot 15 \cdot (-8) = 196 + 480 = 676
$$

Now, substituting into the quadratic formula:
$$
x = \frac{14 \pm \sqrt{676}}{2 \cdot 15}
$$
$$
\sqrt{676} = 26
$$
$$
x = \frac{14 \pm 26}{30}
$$

Calculating the two possible values for $$x$$:
1. $$ x = \frac{40}{30} = \frac{4}{3} $$
2. $$ x = \frac{-12}{30} = -\frac{2}{5} $$

#### Equation 2.1.2: $$ 5^x = \frac{1}{125} $$

We can rewrite $$ \frac{1}{125} $$ as $$ 5^{-3} $$:
$$
5^x = 5^{-3}
$$

Since the bases are the same, we can equate the exponents:
$$
x = -3
$$

### Final Answers:
1. **Irrational**: $$ \sqrt{27} $$
2. **Non-real**: $$ \sqrt{-27} $$
3. **Solutions for $$ 15x^2 - 8 = 14x $$**: $$ x = \frac{4}{3}, -\frac{2}{5} $$
4. **Solution for $$ 5^x = \frac{1}{125} $$**: $$ x = -3 $$
- **Irrational**: $$ \sqrt{27} $$ - **Non-real**: $$ \sqrt{-27} $$ - **Solutions for $$ 15x^2 - 8 = 14x $$**: $$ x = \frac{4}{3}, -\frac{2}{5} $$ - **Solution for $$ 5^x = \frac{1}{125} $$**: $$ x = -3 $$

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