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2. $\sqrt{4x + 13} = x + 2$

Step 1: Isolate the square root. $\sqrt{4x + 13} = x + 2$

Step 2: Square both sides to eliminate the square root. $4x + 13 = (x + 2)^2$

Expand $(x + 2)^2$: $4x + 13 = x^2 + 4x + 4$

Step 3: Simplify and rearrange the terms into standard quadratic form. $0 = x^2 + 4x + 4 - 4x - 13$ $x^2 - 9 = 0$

Step 4: Solve the quadratic equation. $x^2 = 9$ $x = \pm 3$

Step 5: Check for extraneous solutions.

• For $x = 3$: $\sqrt{4(3) + 13} = \sqrt{25} = 5, \quad \text{and } x + 2 = 5.$ Valid solution.

• For $x = -3$: $\sqrt{4(-3) + 13} = \sqrt{1} = 1, \quad \text{but } x + 2 = -1.$ Extraneous solution.

Final Answer: $x = 3$

3. $1 + \sqrt{1 - x} = \sqrt{2x + 4}$

Step 1: Isolate one square root. $\sqrt{1 - x} = \sqrt{2x + 4} - 1$

Step 2: Square both sides. $1 - x = (\sqrt{2x + 4} - 1)^2$

Expand $(\sqrt{2x + 4} - 1)^2$: $1 - x = (2x + 4) - 2\sqrt{2x + 4} + 1$ $1 - x = 2x + 5 - 2\sqrt{2x + 4}$

Step 3: Isolate the remaining square root. $2\sqrt{2x + 4} = 2x + 5 - (1 - x)$ $2\sqrt{2x + 4} = 3x + 4$

Step 4: Square both sides again. $4(2x + 4) = (3x + 4)^2$

Expand both sides: $8x + 16 = 9x^2 + 24x + 16$

Simplify: $0 = 9x^2 + 16x$

Step 5: Factorize. $x(9x + 16) = 0$

Solutions: $x = 0 \quad \text{or} \quad x = -\frac{16}{9}$

Step 6: Check for extraneous solutions. Only $x = 0$ satisfies the original equation.

Final Answer: $x = 0$

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Related Questions:

1. How can you quickly check for extraneous solutions in radical equations?
2. Why do extraneous solutions arise when solving radical equations?
3. How would you solve a radical equation with two square roots on one side?
4. What are the steps for solving a quadratic equation after squaring a radical equation?
5. How can factoring help in solving equations like $9x^2 + 16x = 0$?

Tip:

Always verify your solutions in the original equation to eliminate extraneous solutions caused by squaring!