Respuesta :

Given:

[tex]4x^2+8x+1=0[/tex]

[tex]\begin{gathered} x^2+2x+\frac{1}{4}=0 \\ x^2+2x+1-1+\frac{1}{4}=0 \\ (x+1)^2=1-\frac{1}{4} \\ (x+1)^2=\frac{3}{4} \end{gathered}[/tex]

Taking square root on both sides

[tex]\begin{gathered} x+1=\pm\frac{\sqrt[]{3}}{2} \\ x=-1\pm\frac{\sqrt[]{3}}{2} \\ x=\frac{-2\pm\sqrt[]{3}}{2} \end{gathered}[/tex]

3rd option is the final answer

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