Solve the radical equation. Check all proposed solutions. \( \sqrt{2 x+5}-\sqrt{x-2}=3 \)

To solve the equation \( \sqrt{2x+5}-\sqrt{x-2}=3 \), we can start by isolating one of the square roots. The equation can be rewritten as: \[ \sqrt{2x+5} = \sqrt{x-2} + 3 \] Now, we can square both sides to eliminate the square roots: \[ 2x + 5 = (\sqrt{x-2} + 3)^2 \] Expanding the right side: \[ 2x + 5 = (x - 2) + 6\sqrt{x-2} + 9 \] Simplifying further: \[ 2x + 5 = x + 7 + 6\sqrt{x-2} \] Now, let's get all terms that don't contain the square root on one side: \[ 2x + 5 - x - 7 = 6\sqrt{x-2} \] This simplifies to: \[ x - 2 = 6\sqrt{x-2} \] Next, we isolate the square root again: \[ \sqrt{x-2} = \frac{x-2}{6} \] Square both sides again: \[ x - 2 = \left(\frac{x-2}{6}\right)^2 \] Multiplying each side by \(36\) to eliminate the fraction: \[ 36(x - 2) = (x - 2)^2 \] Expanding both sides: \[ 36x - 72 = x^2 - 4x + 4 \] Rearranging gives us: \[ x^2 - 40x + 76 = 0 \] Next, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -40\), and \(c = 76\): \[ x = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 1 \cdot 76}}{2 \cdot 1} = \frac{40 \pm \sqrt{1600 - 304}}{2} = \frac{40 \pm \sqrt{1296}}{2} \] Calculating the square root: \[ \sqrt{1296} = 36 \] Thus, we have: \[ x = \frac{40 \pm 36}{2} \] This yields two possible solutions: \[ x = \frac{76}{2} = 38 \quad \text{and} \quad x = \frac{4}{2} = 2 \] Finally, we need to check both solutions in the original equation: 1. For \(x = 38\): \[ \sqrt{2(38)+5} - \sqrt{38-2} = \sqrt{76 + 5} - \sqrt{36} = \sqrt{81} - 6 = 9 - 6 = 3 \] This solution works. 2. For \(x = 2\): \[ \sqrt{2(2)+5} - \sqrt{2-2} = \sqrt{4 + 5} - \sqrt{0} = \sqrt{9} - 0 = 3 - 0 = 3 \] This solution also works. Thus, the complete solution set is \(x = 38\) and \(x = 2\).