How do you solve #sqrt[5x-6]=2# and find any extraneous solutions?

1 Answer
Apr 23, 2018

Answer:

#x=2#

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"#