Answer:
D. X=-2+sqrt(3) or -2-sqrt(3)
[tex]x_{1} =-2+\sqrt{3} \\x_{2} =-2-\sqrt{3}[/tex]
Step-by-step explanation:
Quadratic Equation given:
[tex]$\frac{1}{4} x^2+x+\frac{1}{4}=0$[/tex]
In order to get rid of the fractions, multiply both sides by 4.
[tex]$\frac{1}{4}x^2\cdot \:4+x\cdot \:4+\frac{1}{4}\cdot \:4=0\cdot \:4$[/tex]
We get:
[tex]x^2+4x+1=0[/tex]
[tex]x^2+4x+4-4+1=0[/tex]
Completing the square:
[tex](x+2)^2-4+1=0\\[/tex]
[tex](x+2)^2-3=0[/tex]
[tex](x+2)^2=3[/tex]
[tex]x+2=\pm\sqrt{3}[/tex]
[tex]x=-2\pm\sqrt{3}[/tex]
[tex]x_{1} =-2+\sqrt{3} \\x_{2} =-2-\sqrt{3}[/tex]
