Solve the given quadratic equation using the square root property.
$\left(x-\frac12\right)^2-5=0$

• A. $x=\frac12+\sqrt5,x=\frac12-\sqrt5$
• B. $x=\frac52,x=-\frac52$
• C. $x=\frac{\sqrt5}{2},x=-\frac{\sqrt5}{2}$
• D. $x=\frac12,x=-\frac12$

Answer: (A)

Verified step by step guidance

1. Start by isolating the squared term in the equation. The given equation is \((x - \frac{1}{2})^2 - 5 = 0\). Add 5 to both sides to get \((x - \frac{1}{2})^2 = 5\).
2. Apply the square root property, which states that if \(a^2 = b\), then \(a = \pm \sqrt{b}\). Therefore, take the square root of both sides: \(x - \frac{1}{2} = \pm \sqrt{5}\).
3. Solve for \(x\) by adding \(\frac{1}{2}\) to both sides of the equation. This gives \(x = \frac{1}{2} + \sqrt{5}\) and \(x = \frac{1}{2} - \sqrt{5}\).
4. These are the two possible solutions for \(x\) based on the square root property applied to the original equation.
5. Verify the solutions by substituting them back into the original equation to ensure they satisfy \((x - \frac{1}{2})^2 - 5 = 0\).