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

1 Answer
Oct 16, 2015

Answer:

#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.