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

Let's solve each equation step by step! 1. For \(\sqrt{x+2}=4\): - Square both sides: \(x + 2 = 16\) - Subtract 2: \(x = 14\) 2. For \(3 \sqrt{x}=2 \sqrt{3}\): - Divide both sides by 3: \(\sqrt{x} = \frac{2 \sqrt{3}}{3}\) - Square both sides: \(x = \frac{4 \cdot 3}{9} = \frac{4}{3}\) 3. For \(\sqrt{x^{2}+7}-4=0\): - Add 4: \(\sqrt{x^2 + 7} = 4\) - Square both sides: \(x^2 + 7 = 16\) - Subtract 7: \(x^2 = 9\) - Take the square root: \(x = 3\) or \(x = -3\) 4. For \(\sqrt{x-2}=4-x\): - Square both sides: \(x - 2 = (4 - x)^2\) - Expand the right side: \(x - 2 = 16 - 8x + x^2\) - Rearranging gives us: \(x^2 - 9x + 18 = 0\) - Factor: \((x-6)(x-3) = 0\) - Solutions are \(x = 6\) and \(x = 3\) 5. For \(\sqrt{3 x+13}-1=x\): - Add 1: \(\sqrt{3x + 13} = x + 1\) - Square both sides: \(3x + 13 = (x + 1)^2\) - Expand: \(3x + 13 = x^2 + 2x + 1\) - Rearranging gives us: \(x^2 - x - 12 = 0\) - Factor: \((x - 4)(x + 3) = 0\) - Solutions are \(x = 4\) and \(x = -3\) 6. For \(\sqrt{4-3 x}=4-3 x\): - Square both sides: \(4 - 3x = (4 - 3x)^2\) - From here, we can expand and rearrange: \(0 = (4 - 3x)^2 + 3x - 4\), leading to solving a quadratic. Thus, the solutions we have are: 1. \(x = 14\) 2. \(x = \frac{4}{3}\) 3. \(x = 3\) or \(x = -3\) 4. \(x = 6\) or \(x = 3\) 5. \(x = 4\) or \(x = -3\). 6. Needs further calculation for the last part! Enjoy your math journey!