91Ó°ÊÓ

Isolating the square root is the first crucial step in solving equations involving square roots. The goal is to get the square root by itself on one side of the equation. In our exercise, we started with the equation \(\sqrt{4x + 5} - 6 = 2x - 11\). To isolate the square root, we add 6 to both sides, resulting in \(\sqrt{4x + 5} = 2x - 5\). This simplification makes it easier to eliminate the square root in the following steps.
Always remember: whatever you do to one side of the equation, you must do to the other to maintain balance.
Breaking down complex expressions into simpler forms helps in progressing towards the solution systematically.

Once we've isolated the square root term, the next step involves eliminating it by squaring both sides of the equation. Squaring is effective because it removes the square root symbol and turns the expression into a standard polynomial.
In our example, we square both sides of \(\sqrt{4x + 5} = 2x - 5\). This gives us \(\left(\sqrt{4x + 5}\right)^2 = \left(2x - 5\right)^2\), which simplifies to \(4x + 5 = 4x^2 - 20x + 25\).
Eliminating square roots converts the equation into a format we can solve using algebraic techniques, such as factoring.

Factoring quadratic equations is a vital skill when solving polynomials. After eliminating the square root, we rearrange our equation into standard quadratic form \(ax^2 + bx + c = 0\). In our case, simplifying leads to \(0 = 4x^2 - 24x + 20\). We then divide by 4 to get \(x^2 - 6x + 5 = 0\).
Next, we factor the quadratic equation: \(x^2 - 6x + 5 = (x - 1)(x - 5)\). Setting each factor to zero, we find \(x - 1 = 0\) or \(x - 5 = 0\), which gives the solutions \(x = 1\) and \(x = 5\).
The factorization simplifies finding the roots of the equation and bridges the gap between isolating terms and verifying solutions.

Verification of solutions ensures that the solutions derived from the quadratic equation are correct and satisfy the original problem. We substitute each solution back into the original equation.
For \(x = 1\), substituting into the original problem gives \(\sqrt{4(1) + 5} - 6 = 2(1) - 11\), which simplifies to \(\sqrt{9} - 6 = -9\). Since this is not true, \(x = 1\) is not a valid solution.
For \(x = 5\), substituting into the original problem results in \(\sqrt{4(5) + 5} - 6 = 2(5) - 11\), simplifying to \(\sqrt{25} - 6 = -1\). As \(5 - 6 = -1\) holds true, \(x = 5\) is a valid solution.
Verification is essential to confirm that the solutions meet the initial equation's requirements, avoiding errors and ensuring correctness.