Solve the equation: sqrt(2x+3) - sqrt(x+1) = sqrt(x-2)

Fresh day BEN✌️, let's solve. Step 1: Determine the domain of the equation. For the square roots to be defined, the expressions under the radicals must be non-negative: • 2x+3 0 2x -3 x -(3)/(2) • x+1 0 x -1 • x-2 0 x 2 For all three conditions to be met, x must be greater than or equal to 2. So, the domain is x 2. Step 2: Isolate one of the radical terms. The given equation is sqrt(2x+3) - sqrt(x+1) = sqrt(x-2). Move the negative radical term to the right side: sqrt(2x+3) = sqrt(x-2) + sqrt(x+1) Step 3: Square both sides of the equation. (sqrt(2x+3))^2 = (sqrt(x-2) + sqrt(x+1))^2 2x+3 = (x-2) + (x+1) + 2sqrt((x-2)(x+1)) 2x+3 = 2x - 1 + 2sqrt(x^2 + x - 2x - 2) 2x+3 = 2x - 1 + 2sqrt(x^2 - x - 2) Step 4: Isolate the remaining radical term. Subtract 2x-1 from both sides: 2x+3 - (2x-1) = 2sqrt(x^2 - x - 2) 2x+3 - 2x + 1 = 2sqrt(x^2 - x - 2) 4 = 2sqrt(x^2 - x - 2) Step 5: Simplify and square both sides again. Divide both sides by 2: 2 = sqrt(x^2 - x - 2) Square both sides: (2)^2 = (sqrt(x^2 - x - 2))^2 4 = x^2 - x - 2 Step 6: Rearrange into a quadratic equation. x^2 - x - 2 - 4 = 0 x^2 - x - 6 = 0 Step 7: Solve the quadratic equation. Factor the quadratic expression: (x-3)(x+2) = 0 This gives two potential solutions: x-3 = 0 x = 3 x+2 = 0 x = -2 Step 8: Check for extraneous solutions. We must check these solutions against the domain x 2. • For x=3: This value satisfies x 2. Substitute x=3 into the original equation: sqrt(2(3)+3) - sqrt(3+1) = sqrt(3-2) sqrt(9) - sqrt(4) = sqrt(1) 3 - 2 = 1 1 = 1 This solution is valid. • For x=-2: This value does not satisfy x 2. If we substitute x=-2 into sqrt(x-2), we get sqrt(-2-2) = sqrt(-4), which is not a real number. Thus, x=-2 is an extraneous solution. The only valid solution is x=3. x=3 What's next?