The length of a rectangle is 3ft more than twice its width, and the area of the rectangle is 77ft^2, how do you find the dimensions of the rectangle?

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Answer 1 · 2016-03-26T12:17:00

Width = $11/2" ft = 5 foot 6 inches"$

Length = $14" feet"$

Breaking the question down into its component parts:

Let length be $L$
Let width be $w$
Let area be $A$

Length is 3 ft more than: $L=" "?+3$
twice $" "L=2?+3$
its width $" "L=2w+3$

Area $=A = 77="width "xx" Length"$

$A=77= wxx(2w+3)$

$2w^2+3w=77$

$2w^2+3w-77=0$ This is a quadratic equation
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Standard form $y=ax^2+bx+c$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$a= 2"; "b=3"; "c=-77$

$x=(-(3)+-sqrt((-3)^2-4(2)(-77)))/(2(2))$

$x=(-(3)+-25)/4=-7 or 11/2$

As we can not have a negative area in this context the answer for $x$ is $11/2$

But $color(blue)(x=w" so the width is "11/2)$

$color(blue)(L=2w+3 = 11+3=14)$