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the length of a rectangle is 3cm more than twice the width. The area of the rectangle is 90cm^2. Find the dimensions of the rectangle.

Sagot :

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A=l*w
if the length is 3cm more than the width let the length be x+3
and let the width be x
since the area=90cm^2
90=(x+3) *x
90=x(x+3)
90=x²+3x
x²+3x-90=0
x=(-3+ or - √369) /2
x= 8.105 and -11.105
but the dimensions must be positive so they are 8.105 and 11.105 (they are rounded to 3 decimal places, if you don't want them rounded look at them in the
calculator. 
Width= w 
Length= 2w+3
(Length)(Width)= Area

w(2w+3)
2w^2+3w= 90
2w^2+3w-90= 0
(2w+15)(w-6)=0
This can be broken down into two different equations:
2w+15=0 ---------------> w= -15/2
w-6=0 -------------------> w= 6

Since w cannot be negative, the width is 6 cm, and the length is 15 cm. 

This can also be solved using the quadratic formula:
-b±√[b^2-4(a)(c)]
            2a

Start with 2w^2+3w-90= 0
a= 2
b= 3
c= -90
-3±√[9-4(2)(-90)]
           4
-3±√(729)
      4
-3±27
   4
Therefore, the two answers are 6, and -30/4, AKA -15/2.
Once again, 6 is the only one that works because it is positive, so when plugged in, the length is 15, and the width is 6.
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