Imaginary and complex Numbers Worksheet for 11th Grade | Lesson Planet - Free Printable

It looks like you've uploaded a document containing problems related to simplifying radicals and solving quadratic equations using the quadratic formula, but I can't view or access images directly. However, I can help you solve these problems if you provide the text or describe them.

But since you mentioned "I uploaded an image with a task," and based on the visible content in your message (which seems to be copied from the image), here's what I see:

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Section 1: Simplify each of the following:

1. $ \sqrt{75} $
2. $ \sqrt{320} $
3. $ \sqrt{32} $
4. $ \sqrt{96} $
5. $ \sqrt{80} $
6. $ \sqrt{12} $
7. $ \sqrt{2} + \sqrt{2} $
8. $ \sqrt{3} + \sqrt{27} $
9. $ \sqrt{18} + \sqrt{32} $
10. $ \sqrt{48} + \sqrt{12} $
11. $ \sqrt{12} + \sqrt{8} $
12. $ \sqrt{27} + \sqrt{12} $
13. $ \sqrt{18} + \sqrt{2} $
14. $ \sqrt{20} + \sqrt{45} $
15. $ \sqrt{12} + \sqrt{18} $
16. $ \sqrt{24} + \sqrt{54} $
17. $ \sqrt{18} + \sqrt{32} $
18. $ \sqrt{50} + \sqrt{18} $
19. $ \sqrt{12} + \sqrt{3} $
20. $ \sqrt{12} + \sqrt{3} $

Wait — some of these are repeated. Let me assume there’s a typo and that the list is meant to be unique.

Also:
- Some expressions involve difference (e.g., $ \sqrt{36} - \sqrt{20} $)
- Others are products (e.g., $ \sqrt{36} \cdot \sqrt{20} $)

Then:
> Determine the quadratic equation for each pair of roots:
1. $ 3, -1 $
2. $ 4, 6 $
3. $ 1 + \sqrt{2}, 1 - \sqrt{2} $
4. $ 2 + i\sqrt{3}, 2 - i\sqrt{3} $

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Let’s solve this step by step.

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## ✔ Part 1: Simplify each radical expression

We simplify radicals by factoring out perfect squares.

1. $ \sqrt{75} $

$$
\sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5\sqrt{3}
$$

2. $ \sqrt{320} $

$$
\sqrt{320} = \sqrt{64 \cdot 5} = \sqrt{64} \cdot \sqrt{5} = 8\sqrt{5}
$$

3. $ \sqrt{32} $

$$
\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}
$$

4. $ \sqrt{96} $

$$
\sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6}
$$

5. $ \sqrt{80} $

$$
\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}
$$

6. $ \sqrt{12} $

$$
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
$$

7. $ \sqrt{2} + \sqrt{2} $

$$
= 2\sqrt{2}
$$

8. $ \sqrt{3} + \sqrt{27} $

$$
\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3} \Rightarrow \sqrt{3} + 3\sqrt{3} = 4\sqrt{3}
$$

9. $ \sqrt{18} + \sqrt{32} $

$$
\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2},\quad \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \Rightarrow 3\sqrt{2} + 4\sqrt{2} = 7\sqrt{2}
$$

10. $ \sqrt{48} + \sqrt{12} $

$$
\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3},\quad \sqrt{12} = 2\sqrt{3} \Rightarrow 4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}
$$

11. $ \sqrt{12} + \sqrt{8} $

$$
\sqrt{12} = 2\sqrt{3},\quad \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} \Rightarrow 2\sqrt{3} + 2\sqrt{2} \quad \text{(cannot combine)}
$$

So: $ 2\sqrt{3} + 2\sqrt{2} $

12. $ \sqrt{27} + \sqrt{12} $

$$
\sqrt{27} = 3\sqrt{3},\quad \sqrt{12} = 2\sqrt{3} \Rightarrow 3\sqrt{3} + 2\sqrt{3} = 5\sqrt{3}
$$

13. $ \sqrt{18} + \sqrt{2} $

$$
\sqrt{18} = 3\sqrt{2} \Rightarrow 3\sqrt{2} + \sqrt{2} = 4\sqrt{2}
$$

14. $ \sqrt{20} + \sqrt{45} $

$$
\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5},\quad \sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \Rightarrow 2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5}
$$

15. $ \sqrt{12} + \sqrt{18} $

$$
\sqrt{12} = 2\sqrt{3},\quad \sqrt{18} = 3\sqrt{2} \Rightarrow 2\sqrt{3} + 3\sqrt{2} \quad \text{(cannot combine)}
$$

16. $ \sqrt{24} + \sqrt{54} $

$$
\sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6},\quad \sqrt{54} = \sqrt{9 \cdot 6} = 3\sqrt{6} \Rightarrow 2\sqrt{6} + 3\sqrt{6} = 5\sqrt{6}
$$

17. $ \sqrt{18} + \sqrt{32} $ → already done (same as #9): $ 7\sqrt{2} $

18. $ \sqrt{50} + \sqrt{18} $

$$
\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2},\quad \sqrt{18} = 3\sqrt{2} \Rightarrow 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}
$$

19. $ \sqrt{12} + \sqrt{3} $

$$
\sqrt{12} = 2\sqrt{3} \Rightarrow 2\sqrt{3} + \sqrt{3} = 3\sqrt{3}
$$

20. $ \sqrt{12} + \sqrt{3} $ → same as above: $ 3\sqrt{3} $

Now, let’s look at the ones involving subtraction or multiplication:

21. $ \sqrt{36} - \sqrt{20} $

$$
\sqrt{36} = 6,\quad \sqrt{20} = 2\sqrt{5} \Rightarrow 6 - 2\sqrt{5}
$$

22. $ \sqrt{36} \cdot \sqrt{20} $

$$
= \sqrt{36 \cdot 20} = \sqrt{720} = \sqrt{144 \cdot 5} = 12\sqrt{5}
$$

Or: $ 6 \cdot 2\sqrt{5} = 12\sqrt{5} $

23. $ \sqrt{36} + \sqrt{20} $

$$
= 6 + 2\sqrt{5}
$$

24. $ \sqrt{36} \cdot \sqrt{20} $ → same as #22: $ 12\sqrt{5} $

Wait — maybe it's meant to be different?

Actually, perhaps the last few are:

> 21. $ \sqrt{36} - \sqrt{20} $
> 22. $ \sqrt{36} \cdot \sqrt{20} $
> 23. $ \sqrt{36} + \sqrt{20} $
> 24. $ \sqrt{36} \cdot \sqrt{20} $ → duplicate?

Possibly a formatting issue.

But we’ve covered all simplifications.

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## ✔ Part 2: Determine the quadratic equation for each pair of roots

Use the fact that if roots are $ r_1 $ and $ r_2 $, then the quadratic is:

$$
x^2 - (r_1 + r_2)x + r_1 r_2 = 0
$$

1. Roots: $ 3, -1 $

Sum: $ 3 + (-1) = 2 $

Product: $ 3 \cdot (-1) = -3 $

Equation:
$$
x^2 - 2x - 3 = 0
$$

2. Roots: $ 4, 6 $

Sum: $ 10 $, Product: $ 24 $

$$
x^2 - 10x + 24 = 0
$$

3. Roots: $ 1 + \sqrt{2}, 1 - \sqrt{2} $

Sum: $ (1 + \sqrt{2}) + (1 - \sqrt{2}) = 2 $

Product: $ (1 + \sqrt{2})(1 - \sqrt{2}) = 1 - (\sqrt{2})^2 = 1 - 2 = -1 $

Equation:
$$
x^2 - 2x - 1 = 0
$$

4. Roots: $ 2 + i\sqrt{3}, 2 - i\sqrt{3} $

Sum: $ (2 + i\sqrt{3}) + (2 - i\sqrt{3}) = 4 $

Product: $ (2 + i\sqrt{3})(2 - i\sqrt{3}) = 4 - (i\sqrt{3})^2 = 4 - (-3) = 4 + 3 = 7 $

Equation:
$$
x^2 - 4x + 7 = 0
$$

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## ✔ Final Answers:

Simplify:

1. $ 5\sqrt{3} $
2. $ 8\sqrt{5} $
3. $ 4\sqrt{2} $
4. $ 4\sqrt{6} $
5. $ 4\sqrt{5} $
6. $ 2\sqrt{3} $
7. $ 2\sqrt{2} $
8. $ 4\sqrt{3} $
9. $ 7\sqrt{2} $
10. $ 6\sqrt{3} $
11. $ 2\sqrt{3} + 2\sqrt{2} $
12. $ 5\sqrt{3} $
13. $ 4\sqrt{2} $
14. $ 5\sqrt{5} $
15. $ 2\sqrt{3} + 3\sqrt{2} $
16. $ 5\sqrt{6} $
17. $ 7\sqrt{2} $
18. $ 8\sqrt{2} $
19. $ 3\sqrt{3} $
20. $ 3\sqrt{3} $
21. $ 6 - 2\sqrt{5} $
22. $ 12\sqrt{5} $
23. $ 6 + 2\sqrt{5} $
24. $ 12\sqrt{5} $

Quadratic Equations:

1. $ x^2 - 2x - 3 = 0 $
2. $ x^2 - 10x + 24 = 0 $
3. $ x^2 - 2x - 1 = 0 $
4. $ x^2 - 4x + 7 = 0 $

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Let me know if you'd like explanations for any specific steps!