Solve for X

Solve for X

x+1-2sqrt(x+4)=0

\(\begin{array}{|rcll|} \hline x+1-2\sqrt{x+4} &=& 0 \quad & | \quad +2\sqrt{x+4} \\ x+1 &=& 2\sqrt{x+4} \quad & | \quad \text{square both sides} \\ (x+1)^2 &=& 4(x+4) \\ x^2+2x+1 &=& 4x+16 \\ x^2+2x+1 &=& 4x+16 \quad & | \quad -4x \\ x^2+2x-4x+1 &=& 16 \\ x^2-2x+1 &=& 16 \quad & | \quad - 16 \\ x^2-2x+1-16 &=& 0 \\ x^2-2x-15 &=& 0 \\ (x-5)(x+3) &=& 0 \\\\ x_1 = 5 & \text{ or } & x_2 = -3 \\ \hline \end{array} \)

Proof:

\(x_1 = 5:\)
\(\begin{array}{|rcll|} \hline 5+1-2\sqrt{5+4} &\overset{?}{=}& 0 \\ 6-2\sqrt{9} &\overset{?}{=}& 0 \\ 6-2\cdot 3 &\overset{?}{=}& 0 \\ 6-6 &\overset{?}{=}& 0 \\ 0 &\overset{!}{=}& 0 \\\\ x = 5 \text{ is a solution } \\ \hline \end{array} \)

\(x_2 = -3\)
\(\begin{array}{|rcll|} \hline -3+1-2\sqrt{-3+4} &\overset{?}{=}& 0 \\ -2-2\sqrt{1} &\overset{?}{=}& 0 \\ -2-2\cdot 1 &\overset{?}{=}& 0 \\ -2-2 &\overset{?}{=}& 0 \\ -4 &\ne & 0 \\\\ x = -3 \text{ is not a solution } \\ \hline \end{array}\)

The only solution is x = 5