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)   (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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=424 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

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


Answer by Stitch(470)   (Show Source): You can put this solution on YOUR website!
The area of a rectangle is equal to length x width.

Area of original rectangle
Area of original rectangle
The Area of the new rectangle is 3 times larger than the original.

The area of the new rectangle is 105 square meters.
Now lets write the are equation of the new rectangle.

Sub in the given values for L & W

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

Set the equation to zero by subtracting 105 from both sides

Now we can use the quadratic equation
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=424 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

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

-------------------------------------
Since we can only have a positive distance, 4.3m is X.
Add 4.3m to both dimensions to find the new dimensions.




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