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find the dimensions of a rectangle that has a length 3 inches longer than it’s width and has an area of 40in^2

User Dbenham
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1 Answer

10 votes
10 votes

Answer

length: 8 inch and width: 5 inch

Step-by-step explanation

Let width be x,

then the length: x + 3

solve:


\sf (x+3)(x) = 40


\sf x^2+3x = 40


\sf x^2+3x-40=0


\sf x^2+8x-5x-40=0


\sf x(x+8)-5(x+8)=0


\sf (x-5)(x+8)=0


\sf x = 5 \ or \ -8

  • As width cannot be negative, width will be 5.

Find for length:


\sf x + 3


\sf 5+ 3


\sf 8

User Ibn Rushd
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