Question

How do you solve `sqrt(6x-20)=4` and find any extraneous solutions?

Answer 1

See a solution process below:

Explanation:

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

`(sqrt(6x - 20))^2 = 4^2`

`6x - 20 = 16`

Next, add `color(red)(20)` to each side of the equation to isolate the `x` term while keeping the equation balanced:

`6x - 20 + color(red)(20) = 16 + color(red)(20)`

`6x - 0 = 36`

`6x = 36`

Now, divide each side of the equation by `color(red)(6)` to solve for `x` while keeping the equation balanced:

`(6x)/color(red)(6) = 36/color(red)(6)`

`(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 6`

`x = 6`

If we substitute `color(red)(6)` for `color(red)(x)` in the original equation and evaluate the square root we will find the extraneous solutions. Remember, the square root of a number produces a positive and negative result:

`sqrt(6color(red)(x) - 20) = 4` becomes:

`+-sqrt((6 * color(red)(6)) - 20) = 4`

Or

`-sqrt((6 * color(red)(6)) - 20) = 4` and `sqrt((6 * color(red)(6)) - 20) = 4`

`-sqrt(36 - 20) = 4` and `sqrt(36 - 20) = 4`

`-sqrt(16) = 4` and `sqrt(16) = 4`

`-4 != 4` and `4 = 4`

`-sqrt(16)` is an extraneous solution.