# A rectangle has a length pf x . Its width is 3 feet longer than twice the length. Find the dimensions if its a

A rectangle has a length pf x . Its width is 3 feet longer than twice the length. Find the dimensions if its area is 80ft².

(round to the nearst 1/10 )

Length be x

Then width is 2x + 3

Area = Length x Width = 80ft.

80 = x (2x + 3)

= 2x² + 3x

descibe the rest as in qudratic formula

Relevance

80=2x^2+3x

2x^2+3x-80=0

solve for x

x=-3+sqrt(3^2-4(2)(-80)/2(2)

x=5.61 ft

w=2x+3

w=14.24ft

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• enable the width of the rectangle be (x)cm subsequently, length of the rectangle= (3x+6)cm Perimeter of Rectangle= 2(L X B) 76= 2(3x+6) + 2x 6x+12+2x=76 8x=sixty 4 x=8 for this reason, Width=8cm length=(3)(8)+6=30 cm

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• I did not check your set up but here is how to solve:

80 = x (2x + 3)

80 = 2x^2 + 3x

0 = 2x^2 +3x -80

[-3 +- sqrt(3^2 - 4(2)(-80))]/(2(2))

-3 +-sqrt(649)/4

This yields the only positive root of 5.6 for the length.

substitute this value into 2x + 3 to get the width.

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