Question: User: x = 3 + 2 x = 3 - 2
x = 5 or x = 1
(x - 3)^2 = 4
x - 3 = \pm\sqrt{4}
x - 3 = \pm 2
+3 +3
x = 3 \pm 2
6) Simplify
a) i^{47} + i^{93} = i^{140} = (i^{14})^{10} = 1
b) \frac{7}{i} = \frac{7i}{i\cdot i} = \frac{7i}{i^2} = \frac{7i}{-1} = -7i
i^2 = -1, i^3 = -i, i^4 = 1
i^2 \cdot i^4 \cdot i^4 = i^{10}
(-1)(1)(1) = -1
i^{10} \cdot i^4 = i^{14}
(-1)(1) = -1
1. **State the problem:** Simplify the given expressions involving $x$ and powers of $i$.
2. **Solve for $x$:**
Given:
$$ (x - 3)^2 = 4 $$
Take the square root of both sides:
$$ x - 3 = \pm \sqrt{4} $$
$$ x - 3 = \pm 2 $$
Add 3 to both sides:
$$ x = 3 \pm 2 $$
So,
$$ x = 3 + 2 = 5 \quad \text{or} \quad x = 3 - 2 = 1 $$
3. **Simplify powers of $i$:**
Recall the cycle:
$$ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1 $$
and it repeats every 4 powers.
4. **Simplify $i^{47} + i^{93}$:**
Calculate modulo 4:
$$ 47 \mod 4 = 3, \quad 93 \mod 4 = 1 $$
So,
$$ i^{47} = i^3 = -i, \quad i^{93} = i^1 = i $$
Sum:
$$ i^{47} + i^{93} = -i + i = 0 $$
Note: The user's simplification $i^{140} = (i^{14})^{10} = 1$ is incorrect because $i^{47} + i^{93} \neq i^{140}$.
5. **Simplify $\frac{7}{i}$:**
Multiply numerator and denominator by $i$:
$$ \frac{7}{i} = \frac{7i}{i \cdot i} = \frac{7i}{i^2} $$
Since $i^2 = -1$:
$$ \frac{7i}{-1} = -7i $$
6. **Simplify $i^2 \cdot i^4 \cdot i^4$:**
Add exponents:
$$ i^{2 + 4 + 4} = i^{10} $$
Since $i^4 = 1$, $i^{10} = i^{8} \cdot i^{2} = (i^4)^2 \cdot i^2 = 1^2 \cdot (-1) = -1$
7. **Simplify $i^{10} \cdot i^4$:**
Add exponents:
$$ i^{10 + 4} = i^{14} $$
Calculate modulo 4:
$$ 14 \mod 4 = 2 $$
So,
$$ i^{14} = i^2 = -1 $$
**Final answers:**
- $x = 5$ or $x = 1$
- $i^{47} + i^{93} = 0$
- $\frac{7}{i} = -7i$
- $i^2 \cdot i^4 \cdot i^4 = -1$
- $i^{10} \cdot i^4 = -1$