How do you solve sqrtx=6-x?
1 Answer
Explanation:
Since you're dealing with the square root of
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
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