# W is the midpoint of DY. If DW=x^2 +4x and WY=4x+16, how do you find DY?

Oct 27, 2016

Provided the lengths are non-zero
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{\left\mid D Y \right\mid = 64}$

#### Explanation:

If $W$ is the midpoint of $D Y$ then
$\textcolor{w h i t e}{\text{XXX}} \left\mid D W \right\mid = \left\mid W Y \right\mid$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 4 x = 4 x + 16$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 16$

$\textcolor{w h i t e}{\text{XXX}} x = \pm 4$

If $x = - 4$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 4 x = 0 \mathmr{and} 4 x + 16 = 0$
so the total length $\left\mid D Y \right\mid = \left\mid D W \right\mid + \left\mid W Y \right\mid = 0$

If $x = + 4$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 4 x = 32 \mathmr{and} 4 x + 16 = 32$
so the total length $\left\mid D Y \right\mid = \left\mid D W \right\mid + \left\mid W Y \right\mid = 32 + 32 = 64$