SOLUTION: The width of a rectangle is 3 inches less than it's length. The area of the rectangle is 340 square inches. What are the lenght and width of the rectangle?

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Question 28724: The width of a rectangle is 3 inches less than it's length. The area of the rectangle is 340 square inches. What are the lenght and width of the rectangle?
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
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The width of a rectangle is 3 inches less than it's length. The area of the rectangle is 340 square inches. What are the lenght and width of the rectangle?
Let the length of the rectangle be L inches
and let the width of the rectangle be B inches.
The width of a rectangle is 3 inches less than it's length.
That is B =L-3 ----(1)
The area of the rectangle is 340 square inches.
That is (lenth X width) = 340 sqinches
That is (LXB) = 340 ----(2)
Putting (1) in (2) that is substituting for B
L(L-3) = 340
Note: The obvious factors of 340 are 20 and 17 (differing by 3)
But if you are proceeding according to rules,then
L^2-3L = 340
L^2-3L-340 = 0 -----(*)
(L-20)(L+17) =0
L-20 =0 gives L= 20
we don't consider (L+17)=0 giving L=-17 which is negative and length is always positive.
L= 20 in B =L-3 = 20-3 = 17
Therefore Length of the rectangle=20 inches and width = 17 inches
Verification: Area=length X width = 20X17 = 340 which is correct
Note: How did we get (*) as (L-20)(L+17)=0
Hereare the steps:L^2-20L+17L -340 =0
(L^2-20L)+(17L -340) =0
L(L-20)+17(L-20) = 0
Lp+17p = 0 (where p = (L-20) )
p(L+17) = 0
(splitting the middle term (-3L)into two parts so that their sum is the middle term and their product is the product of the square term and the constant term and here since the product (L^2)X(-340)= -340L^2 is negative the two parts should be such that one positive and the other negative and since the middle term is negative, the larger numerical part negative and the other positive and so -340L^2 =-(1X2X2X5X17)L^2 = (-2X2X5 L)X(1X17L) =(-20L)X(17L)