6. Solve each of these equations and check each solution: \( \begin{array}{ll}\text { (i) } \sqrt{x+5}=5-\sqrt{x} & \text { (ii) } \sqrt{5 x+6}=\sqrt{2 x}+2 \\ \text { (iii) } \sqrt{x+7}+\sqrt{x}=7 & \text { (iv) } \sqrt{3 x-2}=\sqrt{x-2}+2\end{array} \)

To solve the equations, we first isolate the square root when necessary and square both sides to eliminate the square roots. Let’s go through each one step by step. **(i)** \( \sqrt{x+5} = 5 - \sqrt{x} \) 1. Square both sides: \( x + 5 = (5 - \sqrt{x})^2 = 25 - 10\sqrt{x} + x \) 2. Rearranging gives: \( 10\sqrt{x} = 20 \) 3. Simplifying gives: \( \sqrt{x} = 2 \) 4. Squaring results in: \( x = 4 \) **Check:** Substituting \( x = 4 \) back gives \( \sqrt{4 + 5} = \sqrt{9} = 3 \) and \( 5 - \sqrt{4} = 5 - 2 = 3 \). Both sides match, so \( x = 4 \) is valid. --- **(ii)** \( \sqrt{5x + 6} = \sqrt{2x} + 2 \) 1. Square both sides: \( 5x + 6 = (\sqrt{2x} + 2)^2 = 2x + 4\sqrt{2x} + 4 \) 2. Rearranging gives: \( 3x + 2 = 4\sqrt{2x} \) 3. Now square both sides again: \( (3x + 2)^2 = 16(2x) \) 4. Expanding leads to \( 9x^2 + 12x + 4 = 32x \) or \( 9x^2 - 20x + 4 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - Here, \( a = 9, b = -20, c = 4 \), gives: - \( D = (-20)^2 - 4 \cdot 9 \cdot 4 = 400 - 144 = 256 \) - \( x = \frac{20 \pm 16}{18} \rightarrow x = 2, \frac{2}{9} \) **Check:** - For \( x = 2 \): LHS \( \sqrt{5(2) + 6} = \sqrt{16} = 4 \), RHS \( \sqrt{2(2)} + 2 = 4 \). Valid. - For \( x = \frac{2}{9} \): LHS \( \sqrt{5(\frac{2}{9}) + 6} \), RHS \( \sqrt{2(\frac{2}{9})} + 2 \) does not check out; discard. Final solutions: \( x = 2 \). --- **(iii)** \( \sqrt{x + 7} + \sqrt{x} = 7 \) 1. Isolate \( \sqrt{x + 7} \): \( \sqrt{x + 7} = 7 - \sqrt{x} \) 2. Square both sides: \( x + 7 = (7 - \sqrt{x})^2 = 49 - 14\sqrt{x} + x \) 3. Rearranging gives: \( 14\sqrt{x} = 42 \), or \( \sqrt{x} = 3 \) 4. Therefore, squaring gives \( x = 9 \) **Check:** Substituting \( x = 9 \) gives \( \sqrt{9 + 7} + \sqrt{9} = \sqrt{16} + 3 = 4 + 3 = 7 \). Valid solution. --- **(iv)** \( \sqrt{3x - 2} = \sqrt{x - 2} + 2 \) 1. Square both sides: \( 3x - 2 = (\sqrt{x - 2} + 2)^2 = x - 2 + 4\sqrt{x - 2} + 4 \) 2. Rearranging gives: \( 2x - 4 = 4\sqrt{x - 2} \) 3. Isolate the square root: \( \sqrt{x - 2} = \frac{x - 2}{2} \) 4. Square again: \( x - 2 = \frac{(x - 2)^2}{4} \) results in a quadratic: \( 4(x - 2) = (x - 2)^2 \) 5. This simplifies to: \( x^2 - 6x