Pre-AP* Algebra II -- A Complete Course - Free Printable

Let's go through each problem in the worksheet and evaluate the expressions using the definition of $ a^{1/n} $, which means:

$$
a^{1/n} = \sqrt[n]{a}
$$

That is, the n-th root of $ a $. We'll also keep in mind that:
- $ a^{1/2} = \sqrt{a} $ (square root)
- $ a^{1/3} = \sqrt[3]{a} $ (cube root), etc.
- For even roots (like square roots), we only consider real numbers when the radicand is non-negative.
- Negative numbers under even roots are not real (they’re imaginary), but cube roots (odd roots) can handle negatives.

---

Problem 1: $ 36^{1/2} $

$$
36^{1/2} = \sqrt{36} = 6
$$

✔ Correct.

---

Problem 2: $ 100^{1/2} $

$$
100^{1/2} = \sqrt{100} = 10
$$

✔ Correct.

---

Problem 3: $ -25^{1/2} $

This is not $ (-25)^{1/2} $, it’s $ -(25^{1/2}) $ because exponentiation takes precedence over the negative sign unless parentheses are used.

So:
$$
-25^{1/2} = -(25^{1/2}) = -\sqrt{25} = -5
$$

✔ Correct.

---

Problem 4: $ (-64)^{1/2} $

$$
(-64)^{1/2} = \sqrt{-64}
$$

This is not a real number, since you cannot take the square root of a negative number in real numbers. The result is imaginary: $ 8i $

But if the question expects real numbers only, then this expression is undefined in real numbers.

✘ Incorrect to write $ \sqrt{-64} $ as a real number.

> So, the answer should be undefined or not a real number.

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Problem 5: $ (-49)^{1/2} $

$$
(-49)^{1/2} = \sqrt{-49}
$$

Again, square root of a negative number → not a real number.

✘ Same issue as #4 — should be undefined in real numbers.

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Problem 6: $ -49^{1/2} $

Same logic as #3:

$$
-49^{1/2} = -(49^{1/2}) = -\sqrt{49} = -7
$$

✔ Correct.

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Problem 7: $ 27^{1/3} $

$$
27^{1/3} = \sqrt[3]{27} = 3
$$

Because $ 3^3 = 27 $

✔ Correct.

---

Problem 8: $ (-64)^{1/3} $

Cube root of a negative number is fine.

$$
(-64)^{1/3} = \sqrt[3]{-64} = -4
$$

Because $ (-4)^3 = -64 $

✔ Correct.

---

Problem 9: $ 81^{1/4} $

$$
81^{1/4} = \sqrt[4]{81}
$$

Note: $ 81 = 3^4 $, so:

$$
\sqrt[4]{81} = \sqrt[4]{3^4} = 3
$$

✔ Correct.

---

Problem 10: $ -32^{1/5} $

Again, this is $ -(32^{1/5}) $, not $ (-32)^{1/5} $

$$
32^{1/5} = \sqrt[5]{32}
$$

Now $ 32 = 2^5 $, so:

$$
\sqrt[5]{32} = 2 \Rightarrow -32^{1/5} = -2
$$

✔ Correct.

---

Problem 11: $ \left( \frac{4}{9} \right)^{1/2} $

$$
\left( \frac{4}{9} \right)^{1/2} = \sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}
$$

✔ Correct.

---

Problem 12: $ \left( \frac{27}{8} \right)^{1/3} $

$$
\left( \frac{27}{8} \right)^{1/3} = \frac{27^{1/3}}{8^{1/3}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2}
$$

✔ Correct.

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## ✔ Summary of Corrections:

| Problem | Student Answer | Correct? | Explanation |
|--------|----------------|---------|------------|
| 1 | 6 | ✔ | $ \sqrt{36} = 6 $ |
| 2 | 10 | ✔ | $ \sqrt{100} = 10 $ |
| 3 | -5 | ✔ | $ -\sqrt{25} = -5 $ |
| 4 | $ \sqrt{-64} $ | ✘ | Not real; undefined in reals |
| 5 | $ \sqrt{-49} $ | ✘ | Not real; undefined in reals |
| 6 | -7 | ✔ | $ -\sqrt{49} = -7 $ |
| 7 | 3 | ✔ | $ \sqrt[3]{27} = 3 $ |
| 8 | -4 | ✔ | $ \sqrt[3]{-64} = -4 $ |
| 9 | 3 | ✔ | $ \sqrt[4]{81} = 3 $ |
| 10 | -2 | ✔ | $ -\sqrt[5]{32} = -2 $ |
| 11 | $ \frac{2}{3} $ | ✔ | $ \sqrt{4/9} = 2/3 $ |
| 12 | $ \frac{3}{2} $ | ✔ | $ \sqrt[3]{27/8} = 3/2 $ |

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## ⚠️ Key Points to Remember:

- $ a^{1/n} = \sqrt[n]{a} $
- For even $ n $ (like 2, 4), $ a $ must be non-negative for real results.
- For odd $ n $ (like 3, 5), negative values are allowed.
- Order of operations: $ -a^{1/n} $ means $ -(a^{1/n}) $, not $ (-a)^{1/n} $
- Only apply the negative sign after evaluating the root.

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Final Answers:

1. $ 6 $
2. $ 10 $
3. $ -5 $
4. Undefined (in real numbers)
5. Undefined (in real numbers)
6. $ -7 $
7. $ 3 $
8. $ -4 $
9. $ 3 $
10. $ -2 $
11. $ \frac{2}{3} $
12. $ \frac{3}{2} $

✔ All answers except 4 and 5 are correct. Problems 4 and 5 involve imaginary numbers and should be marked as undefined in the context of real numbers.