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

Question

Avery Alexander · Accepted Answer

#x=33/2# #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"#

Savannah Anderson · Answer

See a solution process below:

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#

Eva Abramson · Answer

undefined To solve the equation sqrt(2x+3)=6, we need to isolate the variable x.

First, square both sides of the equation to eliminate the square root: (sqrt(2x+3))^2 = 6^2.

Simplifying, we get 2x+3 = 36.

Next, subtract 3 from both sides of the equation: 2x = 36 - 3.

Simplifying further, we have 2x = 33.

Finally, divide both sides of the equation by 2 to solve for x: x = 33/2.

Therefore, the solution to the equation sqrt(2x+3)=6 is x = 33/2.