I really need help on this one please solve.

Question

The length of a rectangle is 4 meters longer than the width. The area of the rectangle is 80 square meters. Find the length and width. Round the answers to the nearest tenth of a meter pleeaase...

Answer 1

Width = x m

Length = (x + 4)m

Area = 80 sqm 

 

(x)m (x+4)m = 80sqm

(x) (x + 4) = 80

x^2 + 4x = 80

x^2 + 4x -80 = 0

 this can not be conveniently factored, so use the quadratic formula

 

a = 1, b = 4, c = -80

 

{-(4) ± √[4^2 - 4(1)(-80)]}/2(1)

[-4 ± √(16+320)]/2

[-4 ± √(336)]/2

 

(-4 ± 4√21)/2 =

-2 ± 2√21

 

Now since we are talking about a physical measurement, it cannot be 

-2 - 2√21 because that would be less than zero, 

 

so, w = -2 + 2√21 

 And l = (-2 + 2√21) + 4 = 2 + 2√21

 

to check the work, multiply (w)(l) = Area

 

(-2 + 2√21)(2 + 2√21) = -4 -4√21 + 4√21 + 84 = -4 + 84 = 80

 

 Converting to decimal w = 7.2 m and l =11.2 m

 

Answer 2

Hello Dusty,

 

Let's set this up.

L = length 

W = width

 

From the problem we know 

L = W+4

and

LW = 80

 

Using substitution

(W+4)W = 80

 

Distributing

W² + 4W = 80

W² + 4W - 80 = 0

 

Use the quadratic formula to solve for W.

 

-b±√(b²-4ac)

---------------  where a,b,c are the coefficients of the quadratic

      2a

 

-4±√[4²-(4)(1)(-80)]       -4±√[16+320]       -4±√336      -4±4√21      

------------------------  =   ----------------  =   ----------- =  ----------  = -2±2√21

             2(1)                           2                       2                  2

 

Since a measurement cannot be negative we eliminate -2-2√21 and 

W = -2+2√21 = 7.2 m

 

Finally, since L = W + 4

L = -2+2√21  + 4

L = 2+2√21 = 11.2 m