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

1 Answer
Mar 12, 2017

Answer:

#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"#