Algebra 12 worksheet focusing on rational exponents and radical expressions.

Explanation:
Let’s go through each problem one by one. The worksheet has two sections: Evaluate and Simplify, all involving radicals and exponents.

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1. Evaluate:

1. $ \sqrt{25} $
Square root of 25 is the number that, when squared, gives 25.
$ 5^2 = 25 $, so $ \sqrt{25} = 5 $

2. $ \sqrt[3]{8} $
Cube root of 8: what number cubed is 8?
$ 2^3 = 8 $, so $ \sqrt[3]{8} = 2 $

3. $ \sqrt{49} $
$ 7^2 = 49 $, so $ \sqrt{49} = 7 $

4. $ \sqrt[3]{27} $
$ 3^3 = 27 $, so $ \sqrt[3]{27} = 3 $

5. $ \sqrt{81} $
$ 9^2 = 81 $, so $ \sqrt{81} = 9 $

6. $ \sqrt[3]{64} $
$ 4^3 = 64 $, so $ \sqrt[3]{64} = 4 $

7. $ \sqrt{100} $
$ 10^2 = 100 $, so $ \sqrt{100} = 10 $

8. $ \sqrt[3]{125} $
$ 5^3 = 125 $, so $ \sqrt[3]{125} = 5 $

✔ So far, all are perfect squares or cubes — easy!

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2. Simplify (write in exponential form):

Remember:
- $ \sqrt[n]{a^m} = a^{m/n} $
- $ \sqrt{a} = a^{1/2} $
- $ \sqrt[3]{a} = a^{1/3} $
- Also, $ (a^m)^n = a^{mn} $
- And $ a^{-m} = \frac{1}{a^m} $

Now simplify:

9. $ x^{1/2} $
Already in exponential form — no change needed.
Answer: $ x^{1/2} $

10. $ x^{2/3} $
Already exponential. Answer: $ x^{2/3} $

11. $ x^{-1/2} $
Already exponential. Answer: $ x^{-1/2} $

12. $ x^{-3/4} $
Already exponential. Answer: $ x^{-3/4} $

13. $ \left(x^2\right)^{1/3} $
Use power rule: $ (x^2)^{1/3} = x^{2 \cdot \frac{1}{3}} = x^{2/3} $

14. $ \left(x^3\right)^{1/2} $
$ = x^{3 \cdot \frac{1}{2}} = x^{3/2} $

15. $ \left(x^{-2}\right)^{1/3} $
$ = x^{-2 \cdot \frac{1}{3}} = x^{-2/3} $

16. $ \left(x^{-3}\right)^{1/2} $
$ = x^{-3 \cdot \frac{1}{2}} = x^{-3/2} $

17. $ \left(\frac{x^2}{y^3}\right)^{1/2} $
Apply exponent to numerator and denominator:
$ = \frac{(x^2)^{1/2}}{(y^3)^{1/2}} = \frac{x^{2 \cdot \frac{1}{2}}}{y^{3 \cdot \frac{1}{2}}} = \frac{x^1}{y^{3/2}} = x y^{-3/2} $
Or keep as $ \frac{x}{y^{3/2}} $ — both are exponential forms. Usually we write with negative exponents if asked for *exponential form*, so:
$ x^1 y^{-3/2} $ → $ x y^{-3/2} $

18. $ \frac{x^{1/2}}{y^{1/3}} $
Already exponential form. Can also write as $ x^{1/2} y^{-1/3} $, which is cleaner exponential form (single expression). So answer: $ x^{1/2} y^{-1/3} $

Let me double-check all:

- 1–8: correct.
- 9–12: already exponential — fine.
- 13: $ (x^2)^{1/3} = x^{2/3} $ ✔
- 14: $ (x^3)^{1/2} = x^{3/2} $ ✔
- 15: $ (x^{-2})^{1/3} = x^{-2/3} $ ✔
- 16: $ (x^{-3})^{1/2} = x^{-3/2} $ ✔
- 17: $ (x^2/y^3)^{1/2} = x^{1} y^{-3/2} $ ✔
- 18: $ x^{1/2}/y^{1/3} = x^{1/2} y^{-1/3} $ ✔

All good.

Final Answer:
1. 5
2. 2
3. 7
4. 3
5. 9
6. 4
7. 10
8. 5
9. $ x^{1/2} $
10. $ x^{2/3} $
11. $ x^{-1/2} $
12. $ x^{-3/4} $
13. $ x^{2/3} $
14. $ x^{3/2} $
15. $ x^{-2/3} $
16. $ x^{-3/2} $
17. $ x y^{-3/2} $
18. $ x^{1/2} y^{-1/3} $