Given:
The eqution is given as, x^2 + 4x + 1 = 0.
The objective is to solve the equation by compleing the square.
Consider the middle of the equation.
![2\cdot a\cdot b=4x](https://img.qammunity.org/2023/formulas/mathematics/college/irtc0ib40bjdalosmdistkayk5art5ttw8.png)
Here, the value of a is x. Then, the value of b can be calculated as,
![\begin{gathered} 2(x)\cdot b=4x \\ b=(4x)/(2x) \\ b=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u08qvhra1aqjdti6vx9bdkrlsv8czfxipo.png)
To complete the equation add +b^2 and -b^2 to the equation.
![\begin{gathered} x^2+4x+2^2-2^2+1=0 \\ x^2+4x+2^2-4+1=0 \\ x^2+4x+2^2-3=0 \\ (x+2)^2-3=0 \\ (x+2)^2=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kqy82zf62b5znhu2e3qhfzoeh6ebrida5f.png)
Take square root on both sides, to solve the value of x,
![\begin{gathered} \sqrt[]{(x+2)^2}=\sqrt[]{3} \\ x+2=\pm\sqrt[]{3} \\ x=\pm\sqrt[]{3}-2 \\ x=+\sqrt[]{3}-2\text{ and -}\sqrt[]{3}-2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3gr7fiqppppv19ycr6jjuysmguzxf8ud2e.png)
Hence, the value of x are +√3-2 and -√3-2.