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

1 Answer
Oct 16, 2015

x=4x=4

Explanation:

Since you're dealing with the square root of xx, you know that xx must be positive, because, for real numbers, you can only take the square root of positive numbers.

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 <= 66x0x6

This means that any potential solution will have to fall in the interval [0, 6][0,6].

Now, square both sides of the equation to get rid of the square root

(sqrt(x))^2 = (6-x)^2(x)2=(6x)2

x = 36 - 12x + x^2x=3612x+x2

Move all the terms on one side of the equation to get

x^2 - 13x + 36 = 0x213x+36=0

Notice that you an rewrite this quadratic as

x^2 - 4x - 9x + 36 = 0x24x9x+36=0

x * (x-4) - 9 * (x-4) = 0x(x4)9(x4)=0

(x-4)(x-9) = 0(x4)(x9)=0

The two solutions will thus be

x-4 = 0 implies x = 4x4=0x=4

and

x - 9 =0 implies x = 9x9=0x=9

Now, because your solutions are restricted to the interval [0, 6][0,6], the only valid solution to the original equation wil be x = 4x=4. The other solution, x = 9x=9, will be extraneous.