# How do you solve x-6 = (x-4)^(1/2) and find any extraneous solutions?

Sep 5, 2016

$x = 8 \mathmr{and} 5$

If $x < 4$ then $x - 4$ is negative $\implies$ solution is in the complex number set of values.

#### Explanation:

Square both sides giving:

${\left(x - 6\right)}^{2} = x - 4$

${x}^{2} - 12 x + 36 = x - 4$

${x}^{2} - 13 x + 40 = 0$

completing the square

$0 = {\left(x - \frac{13}{2}\right)}^{2} + 40 - \frac{169}{4}$

$0 = {\left(x - \frac{13}{2}\right)}^{2} + \frac{160}{4} - \frac{169}{4}$

$0 = {\left(x - \frac{13}{2}\right)}^{2} - \frac{9}{4}$

$x = \frac{13}{2} \pm \sqrt{\frac{9}{4}}$

$x = 8 \mathmr{and} 5$

Sep 5, 2016

$x = 5 \mathmr{and} x = 8$

#### Explanation:

Remember than the index of $\frac{1}{2}$ is the same as the square root.
We cannot find the square root of a negative number, so

$x - 4 \ge 0 \Rightarrow x \ge 4$
$\textcolor{red}{\text{remember to check this later}}$

To get rid of the square root, you need to square both sides. A common error is to square each of the terms of the left, inside of treating them as a binomial.

${\left(x - 6\right)}^{2} = {\left({\left(x - 4\right)}^{\frac{1}{2}}\right)}^{2}$

${x}^{2} - 12 x + 36 = x - 4 \text{ } \leftarrow$ a quadratic, so make = 0.

${x}^{2} - 13 x + 40 = 0 \text{ } \leftarrow$ check for factors.

'Find factors of 40 which add (because of +40) to give 13.'
The signs will be the same, (because of +40) they are both minus (because of -13).

$\text{8 and 5 will work } : 8 \times 5 = 40 \mathmr{and} 8 + 5 = 13$

$\left(x - 5\right) \left(x - 8\right) = 0$

$\text{letting each factor equal 0 gives } x = 5 \mathmr{and} x = 8$

$\textcolor{red}{\text{Both answers are valid}}$