Step-by-Step Solution

Solve the radical equation: $\sqrt{3x + 4} = 2 + \sqrt{2x - 4}$

Step-by-Step Solution

Step 1: Square both sides

$\left(\sqrt{3x + 4}\right)^2 = \left(2 + \sqrt{2x - 4}\right)^2$

$3x + 4 = 4 + 4\sqrt{2x - 4} + (2x - 4)$

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

$x + 4 = 4\sqrt{2x - 4}$

Step 2: Square both sides again

$(x + 4)^2 = \left(4\sqrt{2x - 4}\right)^2$

$x^2 + 8x + 16 = 16(2x - 4)$

$x^2 + 8x + 16 = 32x - 64$

$x^2 - 24x + 80 = 0$

Step 3: Factorize

$(x - 4)(x - 20) = 0$

$x = 4 \quad \text{or} \quad x = 20$

Step 4: Check solutions

• For $x = 4$: $\text{LHS} = \sqrt{3(4) + 4} = \sqrt{16} = 4$ $\text{RHS} = 2 + \sqrt{2(4) - 4} = 2 + 2 = 4 = \text{LHS} \quad \checkmark$

• For $x = 20$: $\text{LHS} = \sqrt{3(20) + 4} = \sqrt{64} = 8$ $\text{RHS} = 2 + \sqrt{2(20) - 4} = 2 + 6 = 8 = \text{LHS} \quad \checkmark$

$\boxed{\text{Solution Set: } \{4, \; 20\}}$