Question

How do you solve `sqrtx=6-x`?

Answer 1

`x=4`

Explanation:

Since you're dealing with the square root of `x`, you know that `x` must be positive, because, for real numbers, you can only take the square root of positive numbers.

`x > =0`

Moreover, you know that the square root of a positive number is always a positive number, so you must have

`6 -x >= 0 implies x <= 6`

This means that any potential solution will have to fall in the interval `[0, 6]`.

Now, square both sides of the equation to get rid of the square root

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

`x = 36 - 12x + x^2`

Move all the terms on one side of the equation to get

`x^2 - 13x + 36 = 0`

Notice that you an rewrite this quadratic as

`x^2 - 4x - 9x + 36 = 0`

`x * (x-4) - 9 * (x-4) = 0`

`(x-4)(x-9) = 0`

The two solutions will thus be

`x-4 = 0 implies x = 4`

and

`x - 9 =0 implies x = 9`

Now, because your solutions are restricted to the interval `[0, 6]`, the only valid solution to the original equation wil be `x = 4`. The other solution, `x = 9`, will be extraneous.