solve $x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{4}{9}$. factor the perfect - square trinomial on the left side…

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. $(x + square)^{2}=\frac{4}{9}$

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. $(x + square)^{2}=\frac{4}{9}$

Answer

Explanation:

Step1: Recall the perfect - square trinomial formula

The perfect - square trinomial formula is (a^{2}+2ab + b^{2}=(a + b)^{2}). For the left - hand side of the equation (x^{2}+\frac{1}{2}x+\frac{1}{16}), we have (a = x). Since (2ab=\frac{1}{2}x) and (a = x), then (2b=\frac{1}{2}), so (b=\frac{1}{4}). And (b^{2}=(\frac{1}{4})^{2}=\frac{1}{16}).

Step2: Factor the left - hand side

So (x^{2}+\frac{1}{2}x+\frac{1}{16}=(x + \frac{1}{4})^{2}).

Answer:

(\frac{1}{4})