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The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle? If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

Respuesta :

Azieq
L = 3 + 2w

119 = w × ( 3 + 2w )
119 = 3w + 2w^2
2w^2 + 3w - 119 = 0
( w - 7 ) ( 2w + 11 ) = 0
w = 7
x = w
x = 7
Hanihan
If x is the width, and the length is 3 more than twice the width,
the length : 2x+3

Area = length x width

(2x+3)x = 119
By expanding equation, [tex]2 x^{2} +3x[/tex] = 119
Thus, [tex]2 x^{2} +3x-119=0[/tex] 
By factorisation, 
(x-7)(2x+17)= 0
x=7 or x= -17/2 (rejected because width cannot be negative)

Thus, the width is 7 and the length is 2(7) +3 = 17

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