Answer
Width: 17
Length: 23
Work Step by Step
Use the equation given in problem 4, but instead of the area as $315$, the area should be $391$
$(2x+11)(2x+17) = 391$
$4x^2+56x+187= 391$
$4x^2 + 56x - 204 = 0$
$4(x^2 + 14x - 51) = 0$
$4(x+17)(x-3) = 0$
Use the zero-product property
$x + 17 = 0$ or $x - 3 = 0$
$x = -17$ or $x = 3$
We take the positive solution since we are talking about area.
Now we need to find the outer dimensions of the frame. This shouldn't be too hard since we know that $(2x+11)$ is the width and $2x + 17$ is the length. Plug in $x = 3$ into each expression and you will find that $17$ is the width and $23$ is the length.
