The length of a rectangle is 2 centimeters less than twice the width. If the area is 84 square centimeters how do you find the dimensions of the rectangle?

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Answer 1 · 2016-11-14T11:21:42

width = 7 cm
length = 12 cm

It is often helpful to draw a quick sketch.

Let length be $L$ $L$
Let width be $w$ $w$

Tony B

Area $= w L$ $=wL$

$= w (2 w - 2)$ $= w(2w-2)$

$= 2 w^{2} - 2 w = 84 c m^{2}$ $= 2w^2-2w" "=" "84 cm^2$
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$Determine w$ $color(blue)("Determine "w)$

Subtract 84 from both sides

$0 = 2 w^{2} - 2 w - 84 \leftarrow this is a quadratic$ $0=2w^2-2w-84" " larr" this is a quadratic"$

I take one look at this and think: 'can not spot how to factorise so use the formula.'

Compare to $y = a x^{2} + b x + c$ $y=ax^2+bx+c" "$ where $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $" "x=(-b+-sqrt(b^2-4ac))/(2a)$

So for our equation we have:

$a = 2; b = - 2; c = - 84$ $a=2"; "b=-2"; "c=-84$

$\Rightarrow w = \frac{2 \pm \sqrt{2^{2} - 4 (2) (- 84)}}{2 (2)}$ $=>w=(2+-sqrt(2^2-4(2)(-84)))/(2(2))$

$w = \frac{2 \pm \sqrt{676}}{4}$ $w=(2+-sqrt(676))/4$

$w = \frac{2}{4} \pm \frac{26}{4}$ $w=2/4+-26/4$

To have $w$ $w$ as a negative value is not logical so go for:

$" "color(green)(ul(bar(|color(white)(.)w=1/2+6 1/2=7 cmcolor(white)(.)|))$

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$color(blue)("Determine "L)$

$L=2w-2$ so substitute for $w$ giving:

$L=2(7)-2 = 12 cm$

$" "color(green)(ul(bar(|color(white)(./.)L=2(7)-2 = 12 cmcolor(white)(.)|))$