91Ó°ÊÓ

Understanding how to isolate the square root in an equation is pivotal when solving radical equations. When you encounter a radical equation like the one in our exercise, \( \sqrt{9x^{2} + 2x - 4} = 3x \), the goal is to have the square root expression by itself on one side of the equation. This provides clarity and makes the next steps of the solution process more straightforward.

In this exercise, the square root is already isolated on the left side, which means there's no requirement for additional steps like moving terms over the equal sign or simplifying the equation before proceeding. This is a crucial initial step as it sets the stage for effectively squaring both sides, which is the next step in solving the equation.

To tackle the radical and expose the variable within, squaring both sides of the equation is necessary. Let's look at our example: \( \sqrt{9x^{2} + 2x - 4} = 3x \). Squaring both sides means raising each side of the equation to the power of two. This is mathematically represented as \( ( \sqrt{9x^{2} + 2x - 4} )^{2} = (3x)^{2} \).

It's critical to square the entire side, not just the terms inside the radical or the terms outside. This removes the square root and simplifies the equation, but remember to apply the exponent correctly. Once squared, our equation becomes \( 9x^{2} + 2x - 4 = 9x^{2} \), leading us to an equation we can solve linearly. Be aware that squaring can introduce extraneous solutions, so it's important to check your answers at the end.

Once the radical has been removed by squaring, we often arrive at a linear equation. Solving it is the final step to finding the value of our variable, which in the exercise is \(x\).

Linear equations are in the form \(ax + b = 0\), and solving them involves isolating \(x\) on one side. In our case, after simplification, we have \(2x - 4 = 0\). To solve for \(x\), we rearrange the equation to isolate the \(x\)-term. Add \(4\) to both sides to get \(2x = 4\), and then divide by \(2\) to find \(x = 2\).

Remember to always check solutions back in the original equation, especially after squaring, to ensure they're not extraneous and actually satisfy the equation. In our solved example, \(x = 2\) checks out when plugged back into the original equation.