How do you solve #sqrt(2x+3)=6#?

2 Answers
Jim G.
Jun 23, 2018

#x=33/2#

Explanation:

#color(blue)"square both sides"#

#(sqrt(2x+3))^2=6^2#

#2x+3=36#

#"subtract 3 from both sides"#

#2x=36-3=33#

#"divide both sides by 2"#

#x=33/2#

#color(blue)"As a check"#

#sqrt(2xx33/2+3)=sqrt(33+3)=sqrt36=6larr"correct"#

smendyka
Jun 23, 2018

See a solution process below:

Explanation:

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

#(sqrt(2x + 3))^2 = 6^2#

#2x + 3 = 36#

Next, subtract #color(red)(3)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x + 3 - color(red)(3) = 36 - color(red)(3)#

#2x + 0 = 33#

#2x = 33#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = 33/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 33/2#

#x = 33/2#

Related questions
See all questions in Radical Equations
Impact of this question
2159 views around the world
You can reuse this answer
Creative Commons License