How do you solve \[\sqrt{2x-3}=5\]?

Hint: This is a question of algebraic equations. The above question can be easily solved by opening the square root and solving for x. These types of questions can also be solved by a hit and trial method where we assume values of x and put it in the equation to verify it.

Complete step by step answer:
Here we are given \[\sqrt{2x-3}=5\]
We will remove the square root of the equation by squaring both sides of the inequality and then solving step by step.
Squaring both sides of the equality does not change the equation or its root, hence we can simply square both sides.
\[\sqrt{2x-3}=5\]
Squaring both sides of the equation
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{2x-3} \right)}^{2}}={{\left( 5 \right)}^{2}}........(1) \\
 & \\
\end{align}\]
Since we know that square of the square root is simply the number without any change that is
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{a} \right)}^{2}}=a \\
 & \\
\end{align}\]
Hence \[{{\left( \sqrt{2x-3} \right)}^{2}}\]= \[2x-3\]
So the equation (1) can be written as
2x-3 = 25
Taking 3 on the other side of the equation we get
\[\Rightarrow \]2x = 25+3
\[\Rightarrow \]2x = 26
Dividing both sides of equality by 2, we get
\[\Rightarrow \]x = \[\dfrac{26}{2}\]
\[\Rightarrow \]x = 13
Hence the solution of the given equation \[\sqrt{2x-3}=5\] is given as x = 13.

Note:
In questions like these where we need to remove the square root we should remember that squaring both sides of the equation does not affect the correct root but it may add additional roots to the equation, for example:
2x – 3 = 5, the solution of this simple equation is x= 4
If we square both sides of the equation we get
\[{{\left( 2x-3 \right)}^{2}}={{5}^{2}}.......(1)\]
Solving the equation (1) we get x= -1 and 4
Whereas the original equation only satisfies x= 4 solution.
Hence in such a case, we must verify our final solution by putting it in the original equation.