SOLUTION: I have been stuck to this question on my math exam..can you solve this for me? Find the dimension of a rectangle with an area of 27 square meters, if the length of a rectangle i

Question 1002096: I have been stuck to this question on my math exam..can you solve this for me?
Find the dimension of a rectangle with an area of 27 square meters, if the length of a rectangle is three more than twice the width.

Found 2 solutions by Theo, Cromlix:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
area of a rectangle = length * width

the length is equal to 3 more than twice the width.

the equation for that is:

length = 2 * width + 3

start with:

area = length * width

replace length with 2 * width + 3 to get:

area = (2 * width + 3) * width

simplify this to get:

area = 2 * width^2 + 3 * width

since the area = 27, then replace area with 27 to get:

27 = 2 * width^2 + 3 * width

subtract 27 from both sides of this equation to get:

0 = 2 * width^2 + 3 * width - 27

this is the same aqs:

2 * width^2 + 3 * width - 27 = 0

this equation is in standard form of ax^2 + bx + c = 0.

replace width with x and your equation becomes:

2x^2 + 3x - 27 = 0

now it's easier to see that your equation is in the standard form of ax^2 + bx + c = 0

since it is in standard form, you can find a and b and c.

a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant term.

in your equation of 2x^2 + 3x - 27 = 0, you get:

a = 2
b = 3
c = -27

don't forget that you made x = width.
you will need that fact later on.

to factor the equation of 2x^2 + 3x - 27 = 0, use the quadratic formula.

the quadratic formula says that:

x = (-b + or - sqrt(b^2-4ac)) / (2a)

this results in 2 possible answers.

x = (-b + sqrt(b^2-4ac))/(2a) or:
x = (-b - sqrt(b^2-4ac))/(2a)

when a = 2 and b = 3 and c = -27, you get:

b^2 = 3^2 = 9

4ac = 4*2*-27) = -216

b^2-4ac = 9-(-216) = 9+216 = 225

sqrt(b^2-4ac) = sqrt(225) = 15

-b = -3

2a = 4

x = (-b + sqrt(b^2-4ac))/(2a) or:
x = (-b - sqrt(b^2-4ac))/(2a) becomes:

x = (-3 + 15)/4 or:
x = (-4 - 15)/4 which becomes:

x = 3 or:
x = -4.5

since x can't be negative, you are left with:

x = 3

now's the time to remember that x represents the width.

go back to your original equation that said:

27 = length * width

since width = 3, this equation becomes 27 = length * 3

solve for length to get length = 9.

you have length = 9 and width = 3

now's the time to confirm your solution satisfies the problem requirements.

those are:

Find the dimension of a rectangle with an area of 27 square meters, if the length of a rectangle is three more than twice the width.

you found the dimensions.

they are length = 9 and width = 3.

an additional requirement is that length is equal to three more than twice the width.

the width is 3.

twice the width is 6

length is 9 which is 3 more than 6.

length is 3 more than twice the width so you have met that requirement.

you're solution is good.

the dimensions are:

length = 9
width = 3

Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
Hi there,
Make the width = 'x'
Length = 3 + 2x
Area of rectangle = Length x width
27 = (3 + 2x)(x)
27 = 3x + 2x^2
2x^2 + 3x - 27 = 0
Factorise:
Using Quadratic formula:

a = 2: b = 3: c = -27:

[[[x = -3+sqrt(225))/4}}} or [[[x = -3-sqrt(225))/4}}}
x = 3 or -4.5(Discount as -ve)
So, Width = 3.0m
Length = 9.0m
Hope this helps :-)
RELATED QUESTIONS
I'm studying for my Maths exam, and I'm stuck on a pythagoras question. Can you solve an... (answered by KMST)
I could use a little help with this math question for my ACT prep worksheets I've been... (answered by lwsshak3)
Please could you help me with a question that I have been set. The question is the... (answered by vleith)
Hello, my name is Rebecca Ghisoiu. I attend Western Govorners University, an online... (answered by Theo)
Hello there, I have been stuck on a math problem for hours and have no idea of how to... (answered by ikleyn)
Hi, folks! Good afternoon! It's my first question here. I've been searchin for a... (answered by Fombitz)
Hello, I'm taking Calculus online and have a homework problem I'm having trouble with. (answered by Alan3354)
I am working with my school website on My Math Lab and this is a question asked. I need... (answered by checkley77,Earlsdon)
I am stuck on a question from my text book, on Inequalities, Permutations, and... (answered by longjonsilver)