[tex]x^2-\frac{3}{4}x+\frac{1}{8}=0[/tex]
0. To complete the square, it is easier to move the constants to the other side:
[tex]x^2-\frac{3}{4}x+\frac{1}{8}-\frac{1}{8}=-\frac{1}{8}[/tex][tex]x^2-\frac{3}{4}x=-\frac{1}{8}[/tex]
2. Getting the square of both sides:
[tex]x^2-\frac{3}{4}x+(\frac{\frac{3}{4}}{2})^2^{}=-\frac{1}{8}+(\frac{\frac{3}{4}}{2})^2[/tex][tex]x^2-\frac{3}{4}x+(\frac{3}{8})^2=-\frac{1}{8}+(\frac{3}{8})^2[/tex][tex]x^2-\frac{3}{4}x+\frac{9}{64}^{}=-\frac{1}{8}+\frac{9}{64}[/tex]
3. Factoring and solving:
[tex](x^{}-\frac{3}{8})^2=\frac{1}{64}[/tex][tex](x^{}-\frac{3}{8})^2-\frac{1}{64}=\frac{1}{64}-\frac{1}{64}[/tex][tex](x^{}-\frac{3}{8})^2-\frac{1}{64}=0[/tex]
Answer:
[tex](x^{}-\frac{3}{8})^2-\frac{1}{64}=0[/tex]
