Solve the equation 2\[0.5ex] vert 2x-5 - 4x-3 = 1

To solve the equation 2sqrt(2x-5) - sqrt(4x-3) = 1: Step 1: Isolate one of the radical terms. 2sqrt(2x-5) = 1 + sqrt(4x-3) Step 2: Square both sides of the equation. (2sqrt(2x-5))^2 = (1 + sqrt(4x-3))^2 4(2x-5) = 1^2 + 2(1)sqrt(4x-3) + (sqrt(4x-3))^2 8x - 20 = 1 + 2sqrt(4x-3) + 4x - 3 8x - 20 = 4x - 2 + 2sqrt(4x-3) Step 3: Isolate the remaining radical term. 8x - 20 - (4x - 2) = 2sqrt(4x-3) 4x - 18 = 2sqrt(4x-3) Divide both sides by 2: 2x - 9 = sqrt(4x-3) Step 4: Square both sides again to eliminate the radical. (2x - 9)^2 = (sqrt(4x-3))^2 4x^2 - 36x + 81 = 4x - 3 Step 5: Rearrange the equation into a standard quadratic form ax^2 + bx + c = 0. 4x^2 - 36x - 4x + 81 + 3 = 0 4x^2 - 40x + 84 = 0 Divide the entire equation by 4 to simplify: x^2 - 10x + 21 = 0 Step 6: Solve the quadratic equation by factoring. We need two numbers that multiply to 21 and add to -10. These numbers are -3 and -7. (x - 3)(x - 7) = 0 This gives two potential solutions: x = 3 or x = 7. Step 7: Check for extraneous solutions by substituting each value back into the original equation. For x = 3: 2sqrt(2(3)-5) - sqrt(4(3)-3) = 1 2sqrt(6-5) - sqrt(12-3) = 1 2sqrt(1) - sqrt(9) = 1 2(1) - 3 = 1 2 - 3 = 1 -1 = 1 This statement is false, so x=3 is an extraneous solution. For x = 7: 2sqrt(2(7)-5) - sqrt(4(7)-3) = 1 2sqrt(14-5) - sqrt(28-3) = 1 2sqrt(9) - sqrt(25) = 1 2(3) - 5 = 1 6 - 5 = 1 1 = 1 This statement is true, so x=7 is the correct solution. The final answer is x=7. Send me the next one 📸