Key Concepts and Formulas

• Solving Radical Equations: Isolate the radical term and square both sides of the equation. Repeat if necessary. Always check for extraneous solutions.
• Binomial Expansion: $(a+b)^2 = a^2 + 2ab + b^2$
• Simplifying Square Roots: $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

Step-by-Step Solution

Step 1: Isolate one of the radical terms.

Given the equation: $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$ We isolate one of the radical terms to prepare for squaring. This makes the next step simpler. We add $\sqrt{2x-1}$ to both sides: $\sqrt{2x + 1} = 1 + \sqrt{2x - 1}$ Explanation: Isolating a radical term before squaring simplifies the process by avoiding squaring a binomial containing two radical terms.

Step 2: Square both sides of the equation.

We square both sides of the equation to eliminate the square root on the left side: $(\sqrt{2x + 1})^2 = (1 + \sqrt{2x - 1})^2$ This simplifies to: $2x + 1 = (1 + \sqrt{2x - 1})^2$ Expanding the right side using the binomial formula $(a+b)^2 = a^2 + 2ab + b^2$, we have: $2x + 1 = 1 + 2\sqrt{2x - 1} + (2x - 1)$ Explanation: Squaring both sides is a crucial step in eliminating the radical. Correct binomial expansion is essential here.

Step 3: Simplify and isolate the remaining radical term.

We simplify the equation and isolate the remaining radical term. $2x + 1 = 1 + 2\sqrt{2x - 1} + 2x - 1$ $2x + 1 = 2x + 2\sqrt{2x - 1}$ Subtracting $2x$ from both sides gives: $1 = 2\sqrt{2x - 1}$ Dividing both sides by 2 isolates the radical: $\sqrt{2x - 1} = \frac{1}{2}$ Explanation: Simplifying and isolating the remaining radical prepares the equation for the next squaring operation.

Step 4: Square both sides again.

We square both sides again to eliminate the remaining square root: $(\sqrt{2x - 1})^2 = \left(\frac{1}{2}\right)^2$ This simplifies to: $2x - 1 = \frac{1}{4}$ Explanation: Squaring both sides again gets rid of the remaining radical, leading to a simple linear equation.

Step 5: Solve for x.

We solve the linear equation for $x$: $2x - 1 = \frac{1}{4}$ Adding 1 to both sides gives: $2x = \frac{1}{4} + 1 = \frac{1}{4} + \frac{4}{4} = \frac{5}{4}$ Dividing both sides by 2 gives: $x = \frac{5}{4} \cdot \frac{1}{2} = \frac{5}{8}$ Explanation: Solving the linear equation gives the potential solution for $x$.

Step 6: Verify the solution.

We must verify the solution $x = \frac{5}{8}$ in the original equation: $\sqrt{2\left(\frac{5}{8}\right) + 1} - \sqrt{2\left(\frac{5}{8}\right) - 1} = 1$ $\sqrt{\frac{5}{4} + 1} - \sqrt{\frac{5}{4} - 1} = 1$ $\sqrt{\frac{9}{4}} - \sqrt{\frac{1}{4}} = 1$ $\frac{3}{2} - \frac{1}{2} = 1$ $\frac{2}{2} = 1$ $1 = 1$ The solution $x = \frac{5}{8}$ is valid. Also, $x = \frac{5}{8} \ge \frac{1}{2}$ so the domain condition is satisfied. Explanation: Verification is crucial to ensure the solution is not extraneous. It confirms the validity of the solution.

Step 7: Substitute the value of x into the expression.

We substitute $x = \frac{5}{8}$ into the expression $\sqrt{4x^2 - 1}$: $\sqrt{4\left(\frac{5}{8}\right)^2 - 1}$ $\sqrt{4\left(\frac{25}{64}\right) - 1}$ $\sqrt{\frac{100}{64} - 1}$ $\sqrt{\frac{25}{16} - 1}$ $\sqrt{\frac{25}{16} - \frac{16}{16}}$ $\sqrt{\frac{9}{16}}$ $\frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}$ Explanation: Substituting the verified value of $x$ into the expression gives the final answer.

Common Mistakes & Tips

• Extraneous Solutions: Always check solutions in the original equation when dealing with radical equations.
• Binomial Expansion: Remember the $2ab$ term when expanding $(a+b)^2$.
• Arithmetic Errors: Double-check calculations, especially when dealing with fractions and square roots.

Summary

We solved the radical equation by isolating the radical terms, squaring both sides, and simplifying until we obtained a linear equation. We found $x = \frac{5}{8}$ and verified that it is a valid solution. Substituting this value into $\sqrt{4x^2 - 1}$ yielded the final result of $\frac{3}{4}$.

Final Answer

The final answer is $\boxed{\frac{3}{4}}$, which corresponds to option (A).