# How do you solve x-12=sqrt(16x)?

Jul 23, 2016

$x = 4 \text{ or } x = 36$

#### Explanation:

What you do to 1 side of the equation you do to the other side.

Lets 'get rid' of the square root.

Square both sides:

${\left(x - 12\right)}^{2} = 16 x$

multiplied out the brackets
${x}^{2} - 24 x + 144 \text{ "=" } 16 x$

Subtract $\textcolor{b l u e}{16 x}$ from both sides
$\textcolor{b r o w n}{{x}^{2} - 24 x \textcolor{b l u e}{- 16 x} + 144 \text{ "=" } 16 x \textcolor{b l u e}{- 16 x}}$

${x}^{2} - 40 x + 144 \text{ "=" } 0$

Note that $\left(- 4\right) \times \left(- 36\right) = + 144 \text{; } - 4 - 36 = 40$

$\left(x - 4\right) \left(x - 36\right) \text{ "=" } 0$

$x = 4 \text{ or } x = 36$
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Check $x - 12 - 4 \sqrt{x} = 0$

,................................................
At $x = 4 \to \left(4\right) - 12 - 4 \left(\textcolor{m a \ge n t a}{\pm 2}\right)$

Try: $\left(4\right) - 12 - 4 \left(\textcolor{m a \ge n t a}{+ 2}\right) = - 16 \textcolor{red}{\text{ Fail!}}$
Try: $\left(4\right) - 12 - 4 \left(\textcolor{m a \ge n t a}{- 2}\right) = 0 \textcolor{g r e e n}{\text{ Works!}}$

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At $x = 36 \to 36 - 12 - 4 \left(\textcolor{m a \ge n t a}{\pm 6}\right)$

Try: $36 - 12 - 4 \left(\textcolor{m a \ge n t a}{+ 6}\right) = 0 \textcolor{g r e e n}{\text{ Works!}}$
Try: $36 - 12 - 4 \left(\textcolor{m a \ge n t a}{- 6}\right) = 48 \textcolor{red}{\text{ Fails!}}$