How do you solve sqrtx=6-x?

1 Answer
Oct 16, 2015

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.