How do you solve #sqrt[5x-6]=2# and find any extraneous solutions?
1 Answer
Apr 23, 2018
Explanation:
#color(blue)"square both sides"#
#"note that "sqrtaxxsqrta=(sqrta)^2=a#
#(sqrt(5x-6))^2=2^2#
#rArr5x-6=4#
#"add 6 to both sides"#
#5xcancel(-6)cancel(+6)=4+6#
#rArr5x=10#
#"divide both sides by 5"#
#(cancel(5) x)/cancel(5)=10/5#
#rArrx=2#
#color(blue)"As a check"# Substitute this value into the left side of the equation and if equal to the right side then it is the solution.
#sqrt(10-6)=sqrt4=2=" right side"#
#rArrx=2" is the solution"#
#"There are no extraneous solutions"#
