\begin{tabular}{l} EXERCISE 15 \\ Olve for \( x \) : \\ \( \begin{array}{l}\sqrt{x+2}=4 \\ 3 \sqrt{x}=2 \sqrt{3} \\ \sqrt{x^{2}+7}-4=0 \\ \sqrt{x-2}=4-x \\ \sqrt{3 x+13}-1=x \\ \sqrt{4-3 x}=4-3 x\end{array} \) \\ \hline\end{tabular}

Sure, let's solve each of the given equations for \( x \) step by step. --- ### 1. \(\sqrt{x + 2} = 4\) **Step 1:** Square both sides to eliminate the square root. \[ (\sqrt{x + 2})^2 = 4^2 \\ x + 2 = 16 \] **Step 2:** Solve for \( x \). \[ x = 16 - 2 \\ x = 14 \] **Solution:** \( x = 14 \) --- ### 2. \(3 \sqrt{x} = 2 \sqrt{3}\) **Step 1:** Isolate the square root. \[ \sqrt{x} = \frac{2 \sqrt{3}}{3} \] **Step 2:** Square both sides to eliminate the square root. \[ (\sqrt{x})^2 = \left(\frac{2 \sqrt{3}}{3}\right)^2 \\ x = \frac{4 \times 3}{9} \\ x = \frac{12}{9} \\ x = \frac{4}{3} \] **Solution:** \( x = \frac{4}{3} \) --- ### 3. \(\sqrt{x^{2} + 7} - 4 = 0\) **Step 1:** Isolate the square root. \[ \sqrt{x^2 + 7} = 4 \] **Step 2:** Square both sides to eliminate the square root. \[ (\sqrt{x^2 + 7})^2 = 4^2 \\ x^2 + 7 = 16 \] **Step 3:** Solve for \( x^2 \). \[ x^2 = 16 - 7 \\ x^2 = 9 \] **Step 4:** Take the square root of both sides. \[ x = \pm 3 \] **Verification:** - For \( x = 3 \): \[ \sqrt{3^2 + 7} - 4 = \sqrt{9 + 7} - 4 = \sqrt{16} - 4 = 4 - 4 = 0 \] - For \( x = -3 \): \[ \sqrt{(-3)^2 + 7} - 4 = \sqrt{9 + 7} - 4 = \sqrt{16} - 4 = 4 - 4 = 0 \] Both solutions are valid. **Solution:** \( x = 3 \) and \( x = -3 \) --- ### 4. \(\sqrt{x - 2} = 4 - x\) **Step 1:** Recognize that both sides must be non-negative. \[ \sqrt{x - 2} \geq 0 \quad \text{and} \quad 4 - x \geq 0 \\ x - 2 \geq 0 \Rightarrow x \geq 2 \\ 4 - x \geq 0 \Rightarrow x \leq 4 \] So, \( 2 \leq x \leq 4 \). **Step 2:** Square both sides to eliminate the square root. \[ (\sqrt{x - 2})^2 = (4 - x)^2 \\ x - 2 = 16 - 8x + x^2 \] **Step 3:** Rearrange the equation. \[ 0 = x^2 - 9x + 18 \] **Step 4:** Solve the quadratic equation. \[ x = \frac{9 \pm \sqrt{81 - 72}}{2} \\ x = \frac{9 \pm \sqrt{9}}{2} \\ x = \frac{9 \pm 3}{2} \] So, \[ x = \frac{12}{2} = 6 \quad \text{or} \quad x = \frac{6}{2} = 3 \] **Step 5:** Check for extraneous solutions within the domain \( 2 \leq x \leq 4 \). - **For \( x = 6 \):** \( 6 \) is not within \( 2 \leq x \leq 4 \). Discard. - **For \( x = 3 \):** \[ \sqrt{3 - 2} = 4 - 3 \\ \sqrt{1} = 1 \\ 1 = 1 \quad \text{(Valid)} \] **Solution:** \( x = 3 \) --- ### 5. \(\sqrt{3x + 13} - 1 = x\) **Step 1:** Isolate the square root. \[ \sqrt{3x + 13} = x + 1 \] **Step 2:** Recognize that the right side must be non-negative. \[ x + 1 \geq 0 \Rightarrow x \geq -1 \] **Step 3:** Square both sides to eliminate the square root. \[ (\sqrt{3x + 13})^2 = (x + 1)^2 \\ 3x + 13 = x^2 + 2x + 1 \] **Step 4:** Rearrange the equation. \[ 0 = x^2 - x - 12 \] **Step 5:** Solve the quadratic equation. \[ x = \frac{1 \pm \sqrt{1 + 48}}{2} \\ x = \frac{1 \pm \sqrt{49}}{2} \\ x = \frac{1 \pm 7}{2} \] So, \[ x = \frac{8}{2} = 4 \quad \text{or} \quad x = \frac{-6}{2} = -3 \] **Step 6:** Check for extraneous solutions within the domain \( x \geq -1 \). - **For \( x = 4 \):** \[ \sqrt{3(4) + 13} - 1 = 4 \\ \sqrt{12 + 13} - 1 = 4 \\ \sqrt{25} - 1 = 4 \\ 5 - 1 = 4 \\ 4 = 4 \quad \text{(Valid)} \] - **For \( x = -3 \):** \( -3 < -1 \). Disregard. **Solution:** \( x = 4 \) --- ### 6. \(\sqrt{4 - 3x} = 4 - 3x\) **Step 1:** Recognize that both sides must be non-negative. \[ \sqrt{4 - 3x} \geq 0 \quad \text{and} \quad 4 - 3x \geq 0 \\ 4 - 3x \geq 0 \Rightarrow x \leq \frac{4}{3} \] Also, the expression under the square root must be non-negative: \[ 4 - 3x \geq 0 \Rightarrow x \leq \frac{4}{3} \] **Step 2:** Let \( y = 4 - 3x \). The equation becomes: \[ \sqrt{y} = y \] **Step 3:** Solve for \( y \). \[ \sqrt{y} = y \\ \sqrt{y} - y = 0 \\ \sqrt{y} (1 - \sqrt{y}) = 0 \] So, \[ \sqrt{y} = 0 \quad \text{or} \quad 1 - \sqrt{y} = 0 \\ y = 0 \quad \text{or} \quad y = 1 \] **Step 4:** Substitute back \( y = 4 - 3x \). - **For \( y = 0 \):** \[ 4 - 3x = 0 \\ 3x = 4 \\ x = \frac{4}{3} \] - **For \( y = 1 \):** \[ 4 - 3x = 1 \\ 3x = 3 \\ x = 1 \] **Step 5:** Verify both solutions within the domain \( x \leq \frac{4}{3} \). - **For \( x = \frac{4}{3} \):** \[ \sqrt{4 - 3\left(\frac{4}{3}\right)} = 4 - 3\left(\frac{4}{3}\right) \\ \sqrt{4 - 4} = 4 - 4 \\ \sqrt{0} = 0 \\ 0 = 0 \quad \text{(Valid)} \] - **For \( x = 1 \):** \[ \sqrt{4 - 3(1)} = 4 - 3(1) \\ \sqrt{4 - 3} = 4 - 3 \\ \sqrt{1} = 1 \\ 1 = 1 \quad \text{(Valid)} \] **Solution:** \( x = \frac{4}{3} \) and \( x = 1 \) --- ## Summary of Solutions 1. \( x = 14 \) 2. \( x = \dfrac{4}{3} \) 3. \( x = 3 \) and \( x = -3 \) 4. \( x = 3 \) 5. \( x = 4 \) 6. \( x = \dfrac{4}{3} \) and \( x = 1 \)