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