# How do you solve sqrtx=6-x?

Oct 16, 2015

$x = 4$

#### 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

$6 - x \ge 0 \implies x \le 6$

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

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

${\left(\sqrt{x}\right)}^{2} = {\left(6 - x\right)}^{2}$

$x = 36 - 12 x + {x}^{2}$

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

${x}^{2} - 13 x + 36 = 0$

Notice that you an rewrite this quadratic as

${x}^{2} - 4 x - 9 x + 36 = 0$

$x \cdot \left(x - 4\right) - 9 \cdot \left(x - 4\right) = 0$

$\left(x - 4\right) \left(x - 9\right) = 0$

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 $\left[0 , 6\right]$, the only valid solution to the original equation wil be $x = 4$. The other solution, $x = 9$, will be extraneous.