Teddy Dillard

2021-12-26

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.

Suhadolahbb

Beginner2021-12-27Added 32 answers

Let the rectangle's weight be w.

length of rectangle be $2w-4$

Area of rectangle =70 sq. feetask

$lenght\times weight=70$

$(2w-4)w=70$

$2{w}^{2}-4w=70$

$2{w}^{2}-4w-70=0$

${w}^{2}-2w-35=0$

${w}^{2}-(7-5)w-35=0$

${w}^{2}-7w+5w-35=0$

$w(w-7)+5(w-7)=0$

$(w-7)(w+5)=0$

$w=7,-5$

$w=-5$ is not possible

$\Rightarrow \text{width of rectangle}=7ft$

$\Rightarrow \text{Length of rectangle}=(2\times 7)-4$

$=14-4$

=10ft

psor32

Beginner2021-12-28Added 33 answers

Let the width be x.

Then the length is$2x-4$ .

Area$=\text{Length}\cdot \text{width}$

$70=x(2x-4)$

$x(x-2)=35$

${x}^{2}-2x-35=0$

${x}^{2}-7x+5x-35=0$

$x(x-7)+5(x-7)=0$

$(x+5)(x-7)=0$

$x=7\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x=-5$

Since the width is positive$x=-5$ is not possible.
$x=7$ is the solution.

The width is 7 feet.

Then the length is

Area

Since the width is positive

The width is 7 feet.

karton

Expert2022-01-04Added 613 answers

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