Answer:
[tex]x=\frac{5}{12}\text{ or }x=-\frac{11}{12}[/tex]
Step-by-step explanation:
[tex]x^2+\frac{1}{2}x+\frac{1}{16}=\frac{4}{9}[/tex]
[tex]x^2+2\times x\times \frac{1}{4}+\frac{1}{16}=\frac{4}{9}[/tex]
[tex](x)^2+2\times x\times \frac{1}{4}+(\frac{1}{4})^2=(\frac{2}{3})^2[/tex]
[tex](x+\frac{1}{4})^2=(\frac{2}{3})^2[/tex] [tex](\because (x+y)^2=x^2+2xy+y^2)[/tex]
[tex]x+\frac{1}{4}=\pm \sqrt{(\frac{2}{3})^2}[/tex]
[tex]x+\frac{1}{4}=\pm (\frac{2}{3})[/tex]
[tex]x=-\frac{1}{4}\pm \frac{2}{3}[/tex]
[tex]x=-\frac{1}{4}+ \frac{2}{3}\text{ or }x=-\frac{1}{4}-\frac{2}{3}[/tex]
[tex]\implies x=\frac{5}{12}\text{ or }x=-\frac{11}{12}[/tex]
