How do you solve # x - 3 = sqrt(x - 1)# and find any extraneous solutions?
1 Answer
Mar 12, 2017
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"#
