How do you solve #5=sqrt(r-3)#?

Question

2211 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-07T15:00:22

See a solution process below:

First, square both sides of the equation to eliminate the radical while keeping the equation balanced:

#5^2 = (sqrt(r - 3))^2#

#25 = r - 3#

Now, add #color(red)(3)# to each side of the equation to solve for #r# while keeping the equation balanced:

#25 + color(red)(3) = r - 3 + color(red)(3)#

#28 = r - 0#

#28 = r#

#r = 28#