Question

How do you solve `sqrt(2x+5) +6=4`?

Answer 1

See a solution process below:

Explanation:

First, subtract `color(red)(6)` from each side of the equation to isolate the radical while keeping the equation balanced:

`sqrt(2x + 5) + 6 - color(red)(6) = 4 - color(red)(6)`

`sqrt(2x + 5) + 0 = -2`

`sqrt(2x + 5) = -2`

Next, square each side of the equation to eliminate the radical while keeping the equation balanced:

`(sqrt(2x + 5))^2 = -2^2`

2x + 5 = 4#

Now, subtract `color(red)(5)` from each side of the equation to isolate the `x` term while keeping the equation balanced:

`2x + 5 - color(red)(5) = 4 - color(red)(5)`

`2x + 0 = -1`

`2x = -1`

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

`(2x)/color(red)(2) = -1/color(red)(2)`

`(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -1/2`

`x = -1/2`

To check the answer we can substitute `-1/2` for `x` and calculate the result:

`sqrt(2x + 5) + 6 = 4` becomes:

`sqrt((2 xx -1/2) + 5) + 6 = 4`

`sqrt(-1 + 5) + 6 = 4`

`sqrt(4) + 6 = 4`

Remember, the square root of a number produces both a positive AND negative result:

`-2 + 6 = 4` and `2 + 6 = 4`

`4 = 4` and `8 != 4`

The solution for `sqrt(4) = 2` is an extraneous solution.

The solution for `sqrt(4) = -2` is a valid solution.