SOLUTION: A rectangular room has an area of 84 square feet. Its width is 9 more than 3 times the length. What are the dimensions of the rectangle?

Question 313097: A rectangular room has an area of 84 square feet. Its width is 9 more than 3 times the length. What are the dimensions of the rectangle?
Found 2 solutions by checkley77, advedantatutor:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
W=3L+9
AREA=LW
84=L(3L+9)
84=3L^9L
3L^2+9L-84=0
3(L^2+3L-28)=0
3(L+7)(L-4)=0
L-4=0
L=4 ANS.
W=3*4+9=12+9=21 ANS.
PROOF:
84=4*21
84=84

Answer by advedantatutor(4) (Show Source): You can put this solution on YOUR website!
Let length = x
let width = y
then x x y = 84 -------------- (1)
But width = 9 more than three times length
therefore width = 3x + 9
y = 3x + 9
Substituting in equation (1) we get
x x(3x + 9) = 84
3x^2 + 9x = 84
3x^2 + 9x -84 = 0
Solving the quadratic equation above we get
x =4, x=-7
Ignoring x = -7 (because length cannot be negative)
we get x = 4
then y = 3x + 9 = 12 + 9 = 21.
Length = 4 feet
Width = 21 feet.
Check:
1) Width = 9 more than 3 times length
= 3 x length + 9
= (3 x 4) + 9
=12 + 9 = 21
2) Area = Length X Width
= 4 x 21 = 84 sq. feet.

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