Given the equation:
[tex]\sqrt[]{x}+1=x-\sqrt[]{x}-1[/tex]
Solving for x:
[tex]\begin{gathered} \sqrt[]{x}+\sqrt[]{x}=x-1-1 \\ 2\sqrt[]{x}=x-2 \end{gathered}[/tex]
Now, we take the square on both sides of the equation:
[tex]\begin{gathered} 4x=x^2-4x+4 \\ 0=x^2-8x+4 \end{gathered}[/tex]
Now, using the general solution of quadratic equations:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
From the problem, we identify:
[tex]\begin{gathered} a=1 \\ b=-8 \\ c=4 \end{gathered}[/tex]
Then, the solutions are:
[tex]\begin{gathered} x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4\cdot1\cdot4}}{2\cdot1}=\frac{8\pm\sqrt[]{64-16}}{2} \\ x=\frac{8\pm4\sqrt[]{3}}{2}=4\pm2\sqrt[]{3} \end{gathered}[/tex]
But the original equation √(x), so x can not be negative if we want a real equation. Then, the only real solution of the equation is:
[tex]x=4+2\sqrt[]{3}[/tex]
