Question:
Solution:
Consider the following expression:
![x-\sqrt[]{x-1}=3](https://img.qammunity.org/2023/formulas/mathematics/college/bf1zozxmshn2gjtu6yu70ggxfr2g06jpfa.png)
solving for the square root, we get:
![x-3=\sqrt[]{x-1}](https://img.qammunity.org/2023/formulas/mathematics/college/fm0zzwckek90s7vg9c1lcwwih6wahpps14.png)
now, squaring both sides of the equation we remove the square root, then we get:

Expanding the left side of the equation, we get:

this is equivalent to:

Solving this quadratic equation by the quadratic formula, we get:

but, notice that for x= 2 we have that:
![2-\sqrt[]{2-1}=2-1=1\text{ }\\e3](https://img.qammunity.org/2023/formulas/mathematics/college/fv86ay1gn9z4mlr6n8010p1jlhu0415arc.png)
and for x= 5 we obtain that:
![5-\sqrt[]{5-1}=5-\sqrt[]{4}\text{ = 5-2 = 3}](https://img.qammunity.org/2023/formulas/mathematics/college/ba5mz3qq6vtbx6i1w7nx4k1kasq9g87jrb.png)
So that, we can conclude that the correct answer is:
x= 5 only, because x=2 is extraneous.