How do you solve #\sqrt { 2x - 1} = 2#?
1 Answer
Dec 2, 2016
Explanation:
Square both sides of the equation.
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(sqrtaxxsqrta=(sqrta)^2=a)color(white)(2/2)|)))#
#rArr(sqrt(2x-1))^2=2^2#
#rArr2x-1=4# add 1 to both sides.
#2xcancel(-1)cancel(+1)=4+1#
#rArr2x=5# To solve for x, divide both sides by 2
#(cancel(2) x)/cancel(2)=5/2#
#rArrx=5/2" is the solution"#
