EXPLORE \& REASON A. Solve $$ 3(a+1)^{2}+2=11 $$. Use at least two different methods. Solving Radical Equations (T) SavvasRealize.com B. Try each of the methods you used in part (a) to solve $$ \sqrt[3]{(a+1)}+2=11 $$. C. Generalize Which of the methods is better suited for solving an equation with a radical? What problems arise when using the other method? HABITS OF MIND Construct Arguments The squares of two numbers are equal. Does that mean that the two numbers themselves must also be equal? Explain. LESSON 5-4 Solving Radical Equations 123

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**A. Solve $$ 3(a+1)^2 + 2 = 11 $$ using two different methods**

**Method 1: Direct Isolation and Square Root**

1. Start with  
   $$
   3(a+1)^2 + 2 = 11.
   $$
2. Subtract $$2$$ from both sides to isolate the term with the square:  
   $$
   3(a+1)^2 = 11 - 2 \quad \Longrightarrow \quad 3(a+1)^2 = 9.
   $$
3. Divide both sides by $$3$$:  
   $$
   (a+1)^2 = \frac{9}{3} \quad \Longrightarrow \quad (a+1)^2 = 3.
   $$
4. Take the square root of both sides (remembering to consider both the positive and negative roots):  
   $$
   a+1 = \pm\sqrt{3}.
   $$
5. Solve for $$a$$:  
   $$
   a = -1 \pm \sqrt{3}.
   $$

**Method 2: Substitution**

1. Let  
   $$
   u = a+1.
   $$
   Then the equation becomes  
   $$
   3u^2 + 2 = 11.
   $$
2. Subtract $$2$$:  
   $$
   3u^2 = 11 - 2 \quad \Longrightarrow \quad 3u^2 = 9.
   $$
3. Divide by $$3$$:  
   $$
   u^2 = 3.
   $$
4. Take the square root:  
   $$
   u = \pm \sqrt{3}.
   $$
5. Substitute back for $$a$$:  
   $$
   a+1 = \pm\sqrt{3} \quad \Longrightarrow \quad a = -1 \pm \sqrt{3}.
   $$

---

**B. Solve $$ \sqrt[3]{(a+1)} + 2 = 11 $$ using each method from part (A)**

**Method 1: Direct Isolation and Cubing**

1. Start with  
   $$
   \sqrt[3]{(a+1)} + 2 = 11.
   $$
2. Subtract $$2$$ from both sides:  
   $$
   \sqrt[3]{(a+1)} = 11 - 2 \quad \Longrightarrow \quad \sqrt[3]{(a+1)} = 9.
   $$
3. Cube both sides to eliminate the cube root:  
   $$
   a+1 = 9^3.
   $$
4. Compute $$9^3$$:  
   $$
   a+1 = 729.
   $$
5. Solve for $$a$$:  
   $$
   a = 729 - 1 \quad \Longrightarrow \quad a = 728.
   $$

**Method 2: Substitution**

1. Let  
   $$
   u = \sqrt[3]{(a+1)}.
   $$
   Then, the equation becomes  
   $$
   u + 2 = 11.
   $$
2. Solve for $$u$$:  
   $$
   u = 11 - 2 \quad \Longrightarrow \quad u = 9.
   $$
3. Since $$u = \sqrt[3]{(a+1)}$$, cube both sides to solve for $$a+1$$:  
   $$
   a+1 = u^3 = 9^3 = 729.
   $$
4. Finally, solve for $$a$$:  
   $$
   a = 729 - 1 \quad \Longrightarrow \quad a = 728.
   $$

---

**C. Generalization: Which Method is Better Suited for Solving an Equation with a Radical?**

When dealing with an equation that involves a radical, the method of isolating the radical and then applying the appropriate power (e.g., cubing for a cube root) is better suited. This approach directly eliminates the radical in one step and minimizes the risk of introducing extraneous solutions.

On the other hand, using the substitution method, while it can work for both squared and other radicals, may be less straightforward if the equation has additional complications. Moreover, if an equation involves an even-index radical (like a square root), isolating the radical and then squaring both sides requires careful checking for extraneous solutions because squaring can introduce solutions that do not satisfy the original equation. For odd-index radicals (like the cube root), this issue is less problematic since the cube function is one-to-one.

---

**Constructing Arguments: Equality of Squares**

If the squares of two numbers are equal, i.e.,  
$$
x^2 = y^2,
$$
this does not necessarily imply that the numbers $$x$$ and $$y$$ themselves are equal. Instead, it implies that  
$$
x = y \quad \text{or} \quad x = -y.
$$
For example, $$3^2 = 9$$ and $$(-3)^2 = 9$$ even though $$3 \neq -3$$.
**A. Solve $$ 3(a+1)^2 + 2 = 11 $$ using two methods** **Method 1:** 1. Subtract 2 from both sides: $$3(a+1)^2 = 9$$. 2. Divide by 3: $$(a+1)^2 = 3$$. 3. Take square root: $$a+1 = \sqrt{3}$$ or $$a+1 = -\sqrt{3}$$. 4. Solve for $$a$$: $$a = -1 + \sqrt{3}$$ or $$a = -1 - \sqrt{3}$$. **Method 2:** 1. Let $$u = a+1$$. 2. Equation becomes $$3u^2 + 2 = 11$$. 3. Subtract 2: $$3u^2 = 9$$. 4. Divide by 3: $$u^2 = 3$$. 5. Take square root: $$u = \sqrt{3}$$ or $$u = -\sqrt{3}$$. 6. Solve for $$a$$: $$a = -1 + \sqrt{3}$$ or $$a = -1 - \sqrt{3}$$. --- **B. Solve $$ \sqrt[3]{(a+1)} + 2 = 11 $$ using both methods** **Method 1:** 1. Subtract 2: $$\sqrt[3]{(a+1)} = 9$$. 2. Cube both sides: $$a+1 = 729$$. 3. Solve for $$a$$: $$a = 728$$. **Method 2:** 1. Let $$u = \sqrt[3]{(a+1)}$$. 2. Equation becomes $$u + 2 = 11$$. 3. Solve for $$u$$: $$u = 9$$. 4. Cube both sides: $$a+1 = 729$$. 5. Solve for $$a$$: $$a = 728$$. --- **C. Generalization:** For equations with even-index radicals (like square roots), isolating the radical and squaring both sides is effective but requires checking for extraneous solutions. For odd-index radicals (like cube roots), isolating and cubing is straightforward and less error-prone. **Constructing Arguments:** If $$x^2 = y^2$$, then $$x = y$$ or $$x = -y$$. For example, $$3^2 = 9$$ and $$(-3)^2 = 9$$, but $$3 \neq -3$$.

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