Question

152. ) Find all real x such that square root x + 1 = x - Square root x - 1.

Answer 1

Given the equation:

`\sqrt[]{x}+1=x-\sqrt[]{x}-1`

Solving for x:

`\begin{gathered} \sqrt[]{x}+\sqrt[]{x}=x-1-1 \\ 2\sqrt[]{x}=x-2 \end{gathered}`

Now, we take the square on both sides of the equation:

`\begin{gathered} 4x=x^2-4x+4 \\ 0=x^2-8x+4 \end{gathered}`

Now, using the general solution of quadratic equations:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

From the problem, we identify:

`\begin{gathered} a=1 \\ b=-8 \\ c=4 \end{gathered}`

Then, the solutions are:

`\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}`

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:

`x=4+2\sqrt[]{3}`