SOLUTION: the length of a rectangle is 4m more than the width.The area is 30m squared. Find the width and length.

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Question 68989: the length of a rectangle is 4m more than the width.The area is 30m squared. Find the width and length.
Found 2 solutions by checkley75, Edwin McCravy:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
L*W=30
L=W+4 thus
(W+4)W=30
W^2+4W=30
W^2+4W-30=0
USING THE QUADRATIC EQUATION WE GET
X=(-B+-SQRT[B^2-4QC])/2A
X=(-4+-SQRT16-4*1*-30])/2*1
X=(-4+-SQRT16+120])/2
X=(-4+-SQRT136)/2
X=(-4+-11.66)/2
X=(-4+11.66)/2
X=7.66/2
X=3.83M IS THE WIDTH.
THUS THE LENGTH IS
L*3.83=30
L=30/3.83
L=7.83M IS THE LENGTH
PROOF
3.83*7.83=30
30=30

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
the length of a rectangle is 4m more than the width.
The area is 30m squared. Find the width and length.

Let the width be x

>>...the length...is 4m more than the width...<<

So the width = x + 4
 ___________
|   x + 4   |
|          x| 
|___________|
  
           A = (length)(width)
          30 = (x + 4)(x)
    x(x + 4) = 30
     x² + 4x = 30
x² + 4x - 30 = 0

That doesn't factor, so we use the quadratic formula:

            x² - 6x + 4 = 0

Use the quadratic formula:
                  ______ 
            -b ± Öb²-4ac
        x = —————————————
                2a 

where a = 1; b = 4; c = -30

                     ______________ 
             -(4) ± Ö(4)²-4(1)(-30)
        x = ————————————————————————
                     2(1) 
                   ______ 
             -4 ± Ö16+120
        x = ——————————————
                  2

                   ___ 
             -4 ± Ö136
        x = ———————————
                 2

                   ____ 
             -4 ± Ö4·34
        x = ————————————
                 2 

                    __
             -4 ± 2Ö34
        x = ———————————
                 2

                      __
             -4     2Ö34
        x = ———— ± ——————
              2      2
                  __
        x = -2 ± Ö34 
                       __
Using the +, x = -2 + Ö34, which
is one solution and equals about 3.830951895
                       __ 
Using the -, x = -2 - Ö34, which
is the other solution and equals about
-7.830951895.
However we discard this since a rectangle's sides 
aren't negative!

So width = x = 3.830951895, and length = x + 4 =
3.830951895 + 4 = 7.830951895

Edwin