SOLUTION: Problem: Originally, a rectangle was three times as long as it is wide. When 2 cm were subtracted from the length and 5 cm were added to the width, the resulting rectangle had an a

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Problem: Originally, a rectangle was three times as long as it is wide. When 2 cm were subtracted from the length and 5 cm were added to the width, the resulting rectangle had an a      Log On


   

Question 251684: Problem: Originally, a rectangle was three times as long as it is wide. When 2 cm were subtracted from the length and 5 cm were added to the width, the resulting rectangle had an area of 90 cm squared. Find the dimensions of the new rectangle.
hmmmm, I tried this problem...but I keep getting decimal answers when solving :/ We are supposed to use only one variable in the actual equation, and use Quadratic equations to solve.
My Attempt:
w=width
l=length
l=3w
Formula-
A=width(length)
(3w-2)(w+5)=90
Then I foiled and got:
3w^2+13w-100=0
And I get stuck there :/ I don't think it factors and I'm not sure if I foiled wrong or set it up wrong. :/ Help me, please? Thanks :)
I put this under Algebra, although I am in Algebra II, I hope that is fine :D
Found 2 solutions by jsmallt9, richwmiller:
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
%283w-2%29%28w%2B5%29=90
Then I foiled and got:
3w%5E2%2B13w-100=0
And I get stuck there :/ I don't think it factors and I'm not sure if I foiled wrong or set it up wrong. :/ Help me, please? Thanks :)
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This is all good. If you think this trinomial does not factor, then use the quadratic formula on it. (This trinomial actually does factor but the quadratic formula still works.)
w+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F2a
Your a = 3, b = 13 and c = -100:
w+=+%28-%2813%29+%2B-+sqrt%28%2813%29%5E2+-+4%283%29%28-100%29%29%29%2F2%283%29
w+=+%28-13+%2B-+sqrt%28169+-+4%283%29%28-100%29%29%29%2F6
w+=+%28-13+%2B-+sqrt%28169+%2B+1200%29%29%2F6
w+=+%28-13+%2B-+sqrt%281369%29%29%2F6
w+=+%28-13+%2B-+37%29%2F6
w+=+%28-13+%2B+37%29%2F6 or w+=+%28-13+-+37%29%2F6
w+=+24%2F6 or w+=+%28-50%29%2F6
w+=+4 or w+=+%28-25%29%2F3
Since w represents the side of a rectangle we will reject the negative solution. So the only possible value for w that works is w = 4. And that makes the original length, which is 3w, 12.

BTW: 3w%5E2%2B13w-100=0 factors into:
%283w+%2B+25%29%28w+-+4%29+=+0
And this gives us the same answers as the quadratic formula did.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
so far so good
It factors to
(w-4)(3w+25) = 0
which yields two answers 4 and -25/3
We need a positive answer.
notice that below that even factored out the 3
and got 3*(w-4)*(w+8.33)
so 3w^2+13w-100=3*(w^2+4 1/3 w-33 1/3)which factors into 3*(w-4)*(w+8 1/3)
You could also have used the quadratic formula. See below
which would have been the best choice after not finding factors.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 3w%5E2%2B13w%2B-100+=+0) has the following solutons:

w%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2813%29%5E2-4%2A3%2A-100=1369.

Discriminant d=1369 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-13%2B-sqrt%28+1369+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%2813%29%2Bsqrt%28+1369+%29%29%2F2%5C3+=+4
w%5B2%5D+=+%28-%2813%29-sqrt%28+1369+%29%29%2F2%5C3+=+-8.33333333333333

Quadratic expression 3w%5E2%2B13w%2B-100 can be factored:
3w%5E2%2B13w%2B-100+=+3%28w-4%29%2A%28w--8.33333333333333%29
Again, the answer is: 4, -8.33333333333333. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B13%2Ax%2B-100+%29

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 1w%5E2%2B4.33333333333333w%2B-33.3333333333333+=+0) has the following solutons:

w%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284.33333333333333%29%5E2-4%2A1%2A-33.3333333333333=152.111111111111.

Discriminant d=152.111111111111 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4.33333333333333%2B-sqrt%28+152.111111111111+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%284.33333333333333%29%2Bsqrt%28+152.111111111111+%29%29%2F2%5C1+=+4


Quadratic expression 1w%5E2%2B4.33333333333333w%2B-33.3333333333333 can be factored:
1w%5E2%2B4.33333333333333w%2B-33.3333333333333+=+1%28w-4%29%2A%28w--8.33333333333333%29
Again, the answer is: 4, -8.33333333333333. Here's your graph: