Question

solve $x^2 + \frac{1}{2}x + \frac{1}{16} = \frac{4}{9}$. factor the perfect - square trinomial on the left side of the equation. $(x + \frac{1}{4})^2 = \frac{4}{9}$ apply the square root property of equality. $x + \quad = \pm \quad$

Explanation:

Step1: Identify factored constant

From the given perfect-square trinomial factorization, the constant inside the parentheses is $\frac{1}{4}$.

Step2: Apply square root property

Take square roots of both sides. The square root of $\frac{4}{9}$ is $\frac{2}{3}$, so we get:
$x + \frac{1}{4} = \pm \frac{2}{3}$

Step3: Solve for x (first case)

Subtract $\frac{1}{4}$ from $\frac{2}{3}$:
$x = \frac{2}{3} - \frac{1}{4} = \frac{8}{12} - \frac{3}{12} = \frac{5}{12}$

Step4: Solve for x (second case)

Subtract $\frac{1}{4}$ from $-\frac{2}{3}$:
$x = -\frac{2}{3} - \frac{1}{4} = -\frac{8}{12} - \frac{3}{12} = -\frac{11}{12}$

Answer:

First blank (for $x + \square$): $\frac{1}{4}$
Second blank (for $\pm \square$): $\frac{2}{3}$
Solutions for $x$: $\frac{5}{12}$ and $-\frac{11}{12}$