# The length of a rectangle is 4 less than twice

The length of a rectangle is 4 less than twice the width. The area of the rectangle is 70 square feet. Find the width, w, of the rectangle algebraically. Explain why one of the solutions for w is not viable.
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Let weigth of rectangle be w
length of rectangle be $2w-4$
Area of rectangle =70 sq. feetask
$lenght×weight=70$
$\left(2w-4\right)w=70$
$2{w}^{2}-4w=70$
$2{w}^{2}-4w-70=0$
${w}^{2}-2w-35=0$
${w}^{2}-\left(7-5\right)w-35=0$
${w}^{2}-7w+5w-35=0$
$w\left(w-7\right)+5\left(w-7\right)=0$
$\left(w-7\right)\left(w+5\right)=0$
$w=7,-5$
$w=-5$ is not possible [width cannot be negtive]
$⇒\text{width of rectangle}=7ft$
$⇒\text{Length of rectangle}=\left(2×7\right)-4$
$=14-4$
=10ft

###### Not exactly what you’re looking for?
psor32
Let the width be x.
Then the length is $2x-4$.
Area $=\text{Length}\cdot \text{width}$
$70=x\left(2x-4\right)$
$x\left(x-2\right)=35$
${x}^{2}-2x-35=0$
${x}^{2}-7x+5x-35=0$
$x\left(x-7\right)+5\left(x-7\right)=0$
$\left(x+5\right)\left(x-7\right)=0$

Since the width is positive $x=-5$ is not possible. $x=7$ is the solution.
The width is 7 feet.
###### Not exactly what you’re looking for?
karton

Explanation:

$w=-5$ isn't viable because measurements have to be above zero.