SOLUTION: The length and width of a rectangle are 7m and 5m, respectively. When each dimension is increased by the same amount, the area is tripled. Find the dimensions of the new rectangle

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Question 929354: The length and width of a rectangle are 7m and 5m, respectively. When each dimension is increased by the same amount, the area is tripled. Find the dimensions of the new rectangle to the nearest tenth of a meter.
Found 2 solutions by ewatrrr, Stitch:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

     .

A = lw

Question States***

(7m + x)(5m + x) = 3(7m)(5m)

35 + 12x + x^2 = 105

x^2 + 12x - 70 = 0 (tossing out negative solution for unit measure)

 x = 4.3 rounded

11.3m  by 9.3m are the dimensions of the new rectangle to the nearest tenth of a meter

 


 
Solved by pluggable solver: SOLVE quadratic equation with variable


Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B12x%2B-70+=+0) has the following solutons:

  

  x%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=%2812%29%5E2-4%2A1%2A-70=424.

  

  Discriminant d=424 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-12%2B-sqrt%28+424+%29%29%2F2%5Ca.

  

    x%5B1%5D+=+%28-%2812%29%2Bsqrt%28+424+%29%29%2F2%5C1+=+4.295630140987

    x%5B2%5D+=+%28-%2812%29-sqrt%28+424+%29%29%2F2%5C1+=+-16.295630140987

    

    Quadratic expression 1x%5E2%2B12x%2B-70 can be factored:

  1x%5E2%2B12x%2B-70+=+1%28x-4.295630140987%29%2A%28x--16.295630140987%29

  Again, the answer is: 4.295630140987, -16.295630140987.
Here's your graph:

graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B12%2Ax%2B-70+%29

Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is equal to length x width.
A+=+L+%2A+W
A+=+7%2A5 Area of original rectangle
highlight%28A+=+35%29 Area of original rectangle
The Area of the new rectangle is 3 times larger than the original.
3+%2A+35+=+105
The area of the new rectangle is 105 square meters.
Now lets write the are equation of the new rectangle.
%28L+%2B+X%29+%2A+%28W+%2B+X%29+=+105
Sub in the given values for L & W
%287+%2B+X%29+%2A+%285+%2B+X%29+=+105
Now we can use foil to simplify the equation
ERROR Algebra::Solver::Engine::invoke_solver_noengine: solver not defined for name 'foil'.
Error occurred executing solver 'foil' .

Now rewrite the equation
X%5E2+%2B+12X+%2B+35+=+105
Set the equation to zero by subtracting 105 from both sides
X%5E2+%2B+12X+-+70+=+0
Now we can use the quadratic equation
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aX%5E2%2BbX%2Bc=0 (in our case 1X%5E2%2B12X%2B-70+=+0) has the following solutons:

X%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=%2812%29%5E2-4%2A1%2A-70=424.

Discriminant d=424 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-12%2B-sqrt%28+424+%29%29%2F2%5Ca.

X%5B1%5D+=+%28-%2812%29%2Bsqrt%28+424+%29%29%2F2%5C1+=+4.295630140987
X%5B2%5D+=+%28-%2812%29-sqrt%28+424+%29%29%2F2%5C1+=+-16.295630140987

Quadratic expression 1X%5E2%2B12X%2B-70 can be factored:
1X%5E2%2B12X%2B-70+=+1%28X-4.295630140987%29%2A%28X--16.295630140987%29
Again, the answer is: 4.295630140987, -16.295630140987. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B12%2Ax%2B-70+%29

-------------------------------------
Since we can only have a positive distance, 4.3m is X.
Add 4.3m to both dimensions to find the new dimensions.
L+=+7+%2B+4.3
highlight_green%28L+=+11.3m%29
W+=+5+%2B+4.3
highlight%28W+=+9.3m%29