Question

How do you solve `\sqrt { 31- 9x } = x - 5`?

Answer 1

Please see the explanation.

Explanation:

The argument for the square root cannot be negative, therefore, we must add the restriction `31 - 9x >= 0` simplified to `x <= 31/9`

`sqrt(31 - 9x) = x - 5; x <= 31/9`

Please notice that the restriction forces us to take only the negative result from the square root operation. If this is unacceptable, then stop here and declare no solution.

Otherwise, square both sides:

`31 - 9x = (x - 5)^2; x <= 31/9`

Expand the square:

`31 - 9x = x^2 - 10x + 25; x <= 31/9`

Add `9x - 31` to both sides:

`0 = x^2 - x - 6;x <= 31/9`

The above factors into:

`0 = (x + 2)(x - 3)`

We dropped the restriction, because both roots are less than `31/9`.

x = -2 and x = 3

checking `x = -2`:

`sqrt(49) = -7`

checking `x = 3`:

`sqrt(4) = -2`

Both checks are true but only, if you allow the result of the square root to be negative.