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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Suhadolahbb · Accepted Answer

Let the rectangle&#039;s weight be w.length of rectangle be&amp;nbsp;2w−4Area of rectangle =70 sq. feetasklenght×weight=70(2w−4)w=702w2−4w=702w2−4w−70=0w2−2w−35=0w2−(7−5)w−35=0w2−7w+5w−35=0w(w−7)+5(w−7)=0(w−7)(w+5)=0w=7,−5w=−5&amp;nbsp;is not possible⇒width of rectangle=7ft⇒Length of rectangle=(2×7)−4=14−4=10ft

psor32 · Accepted Answer

Let the width be x.   Then the length is 2x−4.   Area =Length⋅width   70=x(2x−4)   x(x−2)=35   x2−2x−35=0   x2−7x+5x−35=0   x(x−7)+5(x−7)=0   (x+5)(x−7)=0   x=7 or x=−5   Since the width is positive x=−5 is not possible.  x=7 is the solution.   The width is 7 feet.

karton · Accepted Answer

Explanation: Let w−widthArea=(length)(width)(2w−4)(w)=702w2−4w=70w2−2w=35w2−2w−35=0(w−7)(w+5)=0So w=7 or w=−5 w=−5 isn&#039;t viable because measurements have to be above zero.

The length of a rectangle is 4 less than twice

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