SOLUTION: The area of a rectangle is 45m^2, and the length of the rectangle is 1m more than twice the width. Find the dimensions of the rectangle.

Question 1202194: The area of a rectangle is 45m^2, and the length of the rectangle is 1m more than twice the width. Find the dimensions of the rectangle.

Found 4 solutions by mananth, Theo, ikleyn, josgarithmetic:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
The area of a rectangle is 45m^2, and the length of the rectangle is 1m more than twice the width. Find the dimensions of the rectangle.
Let width be x m
length will be as per the info given 2x+1
Area of rectangle = L * width
x(2x+1) =45
2x^2 +x =45
2x^2+x -45=0
2x^2+10x-9x -45=0
2x(x+5)-9(x+5)=0
(x+5)(2x-9)=0
x=-5 or x= 4.5
x cannot be negative
So x=4.5
Width = 4.5 m
Length = 2x+1 = 2(4.5)+1= 10

10*4.5 =45

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
area of rectangle = 45 square meters.
the length of the rectangle is 1 meter more than twice the width.
L = length
W = width
L = 2W + 1
A = area = 45
A = L * W
replace L with 2W + 1 to get:
A = (2W + 1) * W
simplify to get:
A = 2W^2 + W
since A = 45, equation becomes:
45 = 2W^2 + W
subtract 45 from both sides of the equation to get:
2W^2 + W - 45 = 0
factor to get:
(2W + 10) * (W - 4.5) = 0
solve for W to get:
W = -10 or W = 4.5
W can't be negative, so W = 4.5
since A = 45 and W = 4.5, then L has to be equal to 10.
L = 2W + 1 becomes 10 = 2*4.5 + 1 which becomes 10 = 10, confirming the relationship between L and W is correct.
solution is that the dimensions of the rectangle are 10 by 4.5
10 is the length.
4.5 is the width.
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
The area of a rectangle is 45m^2, and the length of the rectangle is 1m more
than twice the width. Find the dimensions of the rectangle.
~~~~~~~~~~~~~~~~~~~~

        I will show you a simple way to solve the problem MENTALLY.

Let W be the width, in meters; then the length is (2W+1) meters. The area equation is W*(2W+1) = 45. Multiply both sides by 2. You will get 2W*(2W+1) = 90. (1) So, the product of two numbers, (2W) and (2W+1), is 90, and the numbers differ by 1. At this point, you can easily guess the solution: it is 2W = 9 and W = 9/2 = 4.5. Next, left side of equation (1) is monotonically increasing function for positive W. It means that equation (1) has a UNIQUE solution in positive numbers, which we just found. ANSWER. The problem has a unique solution: the width is 4.5 m; the length is 2*4.5+1 = 10 m.

Solved.

-----------------

My reasoning in this solution is mathematically and logically absolutely strict,
although it includes an element of guessing; but this guessing is supported
by another reasoning, which shows that the guessed value is unique.

It is much easier than to factor a quadratic equation on the way
(which you practically will not be able to factor mentally without an error, in any case).

Such approach works for many other similar problems.

////////////////////

        About factoring, I always repeat my standard joke:
        it works especially good, when you know an answer in advance.

Answer by josgarithmetic(39616)   (Show Source): You can put this solution on YOUR website!
w, width
2w+1, length
w(2w+1)=45 for area

Not the only way to solve but

-
discriminant,

RELATED QUESTIONS
The area of a rectangle is 28m^2, and the length of the rectangle is 1m more than twice... (answered by addingup,josgarithmetic)
The length of a rectangle is 1m less than twice its width, and the area of the rectangle... (answered by jim_thompson5910)
The area of a rectangle is 66m^2, and the length of the rectangle is 1m more than double... (answered by josgarithmetic)
The length of a rectangle is 1m more than double the width, and the area of the rectangle (answered by josgarithmetic)
Problem Page The area of a rectangle is 28m^2 , and the length of the rectangle is 1m... (answered by josgarithmetic)
the perimeter of a rectangle is 98m. the length is 1m more than twice the width. Find... (answered by mananth)
The perimeter of a rectangle is 140m. The length is 1m more than twice the width. Find... (answered by JBarnum)
the length of a rectangle is 1m less than four times the width. if the area of the... (answered by ReadingBoosters)
The length of a rectangle is 11 more than twice the width, and the area of the rectangle... (answered by ewatrrr)