Algebra 2 Worksheets | General Functions Worksheets - Free Printable

Let's solve Problem 1 from the worksheet as an example, and then I'll walk you through the method for solving these types of equations. Since you asked to "solve the problem," and there are many problems, I’ll pick Problem 1 and explain how to solve it step-by-step. Then, I’ll briefly summarize the approach for others.

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Problem 1:

$$
\sqrt{x} + 3 = 12
$$

#### Step 1: Isolate the radical
We want to get the square root by itself on one side:
$$
\sqrt{x} = 12 - 3
$$
$$
\sqrt{x} = 9
$$

#### Step 2: Eliminate the square root by squaring both sides
$$
(\sqrt{x})^2 = 9^2
$$
$$
x = 81
$$

#### Step 3: Check your solution
Plug $ x = 81 $ back into the original equation:
$$
\sqrt{81} + 3 = 9 + 3 = 12 \quad \text{✓}
$$
✔ The solution checks out.

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✔ Final Answer for Problem 1:

$$
\boxed{x = 81}
$$

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Now let’s go over how to solve other similar problems in this set.

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## 🔍 General Strategy for Solving Radical Equations:

1. Isolate the radical expression (square root, cube root, etc.) on one side.
2. Raise both sides to the power that eliminates the root:
- For square roots: square both sides.
- For cube roots: cube both sides.
- For $ x^{1/2} $: square both sides.
- For $ x^{1/3} $: cube both sides.
3. Solve the resulting equation.
4. Check for extraneous solutions — especially when dealing with even roots (like square roots), because squaring can introduce false solutions.

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Let’s do another one: Problem 5

Problem 5:

$$
5\sqrt{3x} = 15
$$

#### Step 1: Divide both sides by 5
$$
\sqrt{3x} = 3
$$

#### Step 2: Square both sides
$$
(\sqrt{3x})^2 = 3^2
\Rightarrow 3x = 9
$$

#### Step 3: Solve for $ x $
$$
x = 3
$$

#### Step 4: Check
$$
5\sqrt{3(3)} = 5\sqrt{9} = 5 \cdot 3 = 15 \quad \text{✓}
$$

✔ Solution: $ \boxed{x = 3} $

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Problem 10:

$$
\sqrt[3]{x} + 1 = -2
$$

#### Step 1: Isolate cube root
$$
\sqrt[3]{x} = -3
$$

#### Step 2: Cube both sides
$$
x = (-3)^3 = -27
$$

#### Step 3: Check
$$
\sqrt[3]{-27} + 1 = -3 + 1 = -2 \quad \text{✓}
$$

✔ Solution: $ \boxed{x = -27} $

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Problem 19:

$$
x^{5/3} = 243
$$

This means: $ x^{5/3} = 243 $

To eliminate the exponent, raise both sides to the reciprocal power: $ \frac{3}{5} $

$$
(x^{5/3})^{3/5} = 243^{3/5}
\Rightarrow x = 243^{3/5}
$$

Now simplify $ 243^{3/5} $. Note: $ 243 = 3^5 $

So:
$$
243^{3/5} = (3^5)^{3/5} = 3^{5 \cdot (3/5)} = 3^3 = 27
$$

✔ Solution: $ \boxed{x = 27} $

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Problem 25:

$$
\sqrt{x - 3} = \sqrt{2x - 7}
$$

#### Step 1: Square both sides
$$
(\sqrt{x - 3})^2 = (\sqrt{2x - 7})^2
\Rightarrow x - 3 = 2x - 7
$$

#### Step 2: Solve
$$
x - 3 = 2x - 7 \\
-3 + 7 = 2x - x \\
4 = x
$$

#### Step 3: Check
Left: $ \sqrt{4 - 3} = \sqrt{1} = 1 $
Right: $ \sqrt{2(4) - 7} = \sqrt{8 - 7} = \sqrt{1} = 1 $ ✔

✔ Solution: $ \boxed{x = 4} $

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Problem 33: Velocity on Earth

Given:
$ V = \sqrt{2gh} $, $ V = 30 $ m/s, $ g = 9.81 $ m/s²

Solve for $ h $:

$$
30 = \sqrt{2 \cdot 9.81 \cdot h}
$$

Square both sides:
$$
900 = 2 \cdot 9.81 \cdot h = 19.62h
$$

$$
h = \frac{900}{19.62} \approx 45.87 \text{ meters}
$$

✔ Height: $ \boxed{h \approx 45.87} $ meters

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Summary of Key Concepts:

| Type of Equation | Strategy |
|------------------|---------|
| $ \sqrt{x} = a $ | Square both sides |
| $ \sqrt[3]{x} = a $ | Cube both sides |
| $ x^{m/n} = a $ | Raise both sides to $ n/m $ |
| $ \sqrt{A} = \sqrt{B} $ | Square both sides, check domain |
| Rational exponents | Use reciprocal powers |

Always check for extraneous solutions, especially when squaring both sides.

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If you'd like me to solve all the problems or focus on specific ones (e.g., #33–35), just let me know!