Question

How do you solve `x - 3 = sqrt(x - 1)` and find any extraneous solutions?

Answer 1

`x=5," extraneous solution " x=2`

Explanation:

`color(blue)"square both sides"`

`(x-3)^2=(sqrt(x-1))^2`

`rArrx^2-6x+9=x-1`

Equate the quadratic to zero

`rArrx^2-7x+10=0`

Factorising the quadratic gives.

`(x-2)(x-5)=0`

`rArrx=5" or "x=2`

`color(blue)"As a check"`

Substitute these values into the equation and if the left side equals the right side then they are the solutions.

`• x=5to5-3=sqrt(5-1)to2=2rArr" a solution"`

`• x=2to2-3=sqrt1to-1=1rArr" not a solution"`

`rArrx=5" is the solution, but "x=2" is extraneous"`