Question

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

Answer 1

`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"`

Answer 2

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`