A rectangle has a length that is 3 more than the width. If the area of the rectangle is 28 units squared, what are the dimensions of the rectangle?​

you can think this

WL = 28

L=3w + 8

W( 3W +8) = 28

3 W^2 + 8W + = 28

3W^2 + 8W -28 = 0 (1)

3 W^2 - 6 W + !4 W -28 =0-

3 W ( W - 2 ) + 14 ( W -2) =0

(3W+ 14) ( w-2) = 0

w = 2 L = 3(2) + 8 = 14 , then LW =28

ignore the negative answer of w = -14/3

If we write all factors of 28 { 2, 4 , 7, 14)

We see that 28 = 2 * 14 , 14 = 2* 3 + 8

Therefore we can have a rectangle with area of 28 units, with 2, 14 be its dimension , and have the

property of L = 2*3 +8 (

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Nicole B. answered • 01/18/14

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Hi Victoria,

Area=Length x width= 28 sq in

L=8+3w

So substitute that back into the equation:

Area = L x W

28 sq in = (8+3w) x w

28 sq in = 8w +3w2

28 sq in = w (8 + 3w)

Possible factors of 28 are 2 and 14 or 4 and 7, and 1 and 28. Substituting the possibilities in the equation, we find that the correct set of factors are 2 and (8 + 3*2) or 14.

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Arthur D. answered • 01/18/14

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w=width

l=3w+8

A=l*w

28=(3w+8)(w)

28=3w^2+8w

3w^2+8w-28=0

(3w+14)(w-2)=0

3w+14=0

3w=-14

w=-14/3, disregard this answer because it is negative

w-2=0

w=2

l=3*2+8

l=14

check: 28=14*2=28

l=14 inches

w=2 inches

14 is 8 more than 3 times 2 !

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Tom D. answered • 01/18/14

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1) A=LW = 28 in

2) L = 3W + 8

Substitute L = 28/W into the 2nd equation

28/W = 3W + 8

28 = 3W^2 + 8W

3W^2 + 8W - 28 = 0

(3W+ 14)(W - 2) = 0

Note that the answer must be positive therefore the second 'zero' is the answer