How do you solve sqrtx=6-x√x=6−x?
1 Answer
Explanation:
Since you're dealing with the square root of
x > =0x>=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 <= 66−x≥0⇒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)2=(6−x)2
x = 36 - 12x + x^2x=36−12x+x2
Move all the terms on one side of the equation to get
x^2 - 13x + 36 = 0x2−13x+36=0
Notice that you an rewrite this quadratic as
x^2 - 4x - 9x + 36 = 0x2−4x−9x+36=0
x * (x-4) - 9 * (x-4) = 0x⋅(x−4)−9⋅(x−4)=0
(x-4)(x-9) = 0(x−4)(x−9)=0
The two solutions will thus be
x-4 = 0 implies x = 4x−4=0⇒x=4
and
x - 9 =0 implies x = 9x−9=0⇒x=9
Now, because your solutions are restricted to the interval