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When working with equations involving square roots, the first step is to 'isolate the square root'. But what does this mean? Isolating the square root refers to moving all other terms away from the square root term to the other side of the equation. This is a crucial step because it simplifies the equation, making it easier to work with.

In our example, the equation starts as \(\sqrt{4y} + 3 = 5\). To isolate \(\sqrt{4y}\), we need to move the '+3' to the other side by subtracting 3 from both sides, which results in \(\sqrt{4y} = 2\). Isolating the square root sets us up for the next important step, which is squaring both sides to eliminate the square root.

After isolating the square root, the next pivotal step is 'squaring both sides' of the equation. This involves raising both sides of the equation to the power of two. The reason for doing this is to remove the square root, as squaring the square root and the number inside it turns it back into a simple algebraic expression without a radical.

Let's apply this to the isolated square root equation we have, \(\sqrt{4y} = 2\). By squaring both sides, we get \(\sqrt{4y}^2 = 2^2\), which simplifies to \(4y = 4\). This squared equation is now much easier to solve for the variable \(y\). Squaring both sides is a vital transformation and must be handled with care to ensure the equation is balanced and the integrity of the solution is maintained.

After manipulating an algebraic equation, especially one with a square root, it's essential to 'verify the solutions'. This step is the final quality check to ensure that the solutions we've found satisfy the original equation. It is possible to introduce extraneous solutions when squaring both sides, because squaring is not a one-to-one function—in other words, it doesn’t always go back to the original value uniquely. Therefore, placing our found value of \(y\) back into the original equation is crucial.

For our example, we found that \(y = 1\). To verify, we substitute \(y = 1\) into the initial equation \(\sqrt{4y} + 3 = 5\), yielding \(\sqrt{4(1)} + 3 = 5\), which simplifies to the correct equation, \(2 + 3 = 5\). Since both sides match, we confirm that \(y = 1\) is indeed a valid solution and our answer checks out.