Question:
Solution:
Consider the following expression:
[tex]x-\sqrt[]{x-1}=3[/tex]
solving for the square root, we get:
[tex]x-3=\sqrt[]{x-1}[/tex]
now, squaring both sides of the equation we remove the square root, then we get:
[tex](x-3)^2=x-1[/tex]
Expanding the left side of the equation, we get:
[tex]x^2-6x+9^{}=x-1[/tex]
this is equivalent to:
[tex]x^2-7x+10^{}=0[/tex]
Solving this quadratic equation by the quadratic formula, we get:
[tex]x=\text{ 5 or x=2}[/tex]
but, notice that for x= 2 we have that:
[tex]2-\sqrt[]{2-1}=2-1=1\text{ }\ne3[/tex]
and for x= 5 we obtain that:
[tex]5-\sqrt[]{5-1}=5-\sqrt[]{4}\text{ = 5-2 = 3}[/tex]
So that, we can conclude that the correct answer is:
x= 5 only, because x=2 is extraneous.
