Solve 3 times the square root of (2x - 1) equals 14.

Step-by-step solution

1. 1

   Isolate the radical before squaring. Squaring $3\sqrt{2x-1}$ directly would give $9(2x-1)$, which works but invites arithmetic slips. Divide by the coefficient first: $\sqrt{2x-1} = \frac{14}{3}.$

2. 2

   Note that no extraneous roots can appear. The right-hand side $\tfrac{14}{3}$ is a positive constant, so squaring is reversible here and every solution of the squared equation will be genuine.

3. 3

   Square both sides. $2x-1 = \left(\frac{14}{3}\right)^2 = \frac{196}{9}.$

4. 4

   Add 1 using a common denominator. $2x = \frac{196}{9}+\frac99 = \frac{205}{9}.$

5. 5

   Divide by 2. $x = \frac{205}{18} \approx 11.3889.$ Since $205$ and $18$ share no factor, this is already in lowest terms.

6. 6

   Check the domain and the answer. $2x-1 = \tfrac{205}{9}-1 = \tfrac{196}{9} > 0$, so the radicand is valid, and $3\sqrt{\tfrac{196}{9}} = 3\cdot\tfrac{14}{3} = 14$ exactly as required.