Answer:
For a rectangle of length L and width W, the area is:
A = L*W
In this case, we start with:
L = 14ft
W = 10ft
Then the initial area is:
A = 14ft*10ft = 140ft^2
a) Now she wants to increase the same amount in each measure, then the new measures will be:
L = 14ft + x
W = 10ft + x
Now we want this new area to be twice the one that we got before, then the equation will be:
(14ft + x)*(10ft + x) = 2*140ft^2 = 280ft^2.
b) Let's solve the equation:
(14ft + x)*(10ft + x) = 280ft^2
First lets distribute the multiplication:
140ft + x*10ft + 14ft*x + x^2 = 280ft
Now let's group terms with the same power of x.
x^2 + (10ft + 14ft)*x + 140ft - 280ft = 0.
x^2 + 24ft*x - 140ft = 0.
Now we need to solve this quadratic equation, here we must use the Bhaskara equation for quadratic equations, and we will get two solutions for x.
[tex]x = \frac{-24ft +- \sqrt{(24ft)^2 - 4*140ft*1} }{2*1} = \frac{-24ft +- 33.7ft}{2}[/tex]
Then the two solutions are:
x = (-24ft - 33.7ft)/2 = -28.9 ft
This solution is negative, so it does not make sense and we can discard it.
x = (-24ft + 33.7ft)/2 = 4.85ft
This is positive, so we can use this.
Now we can find the new length and width:
L = 14ft + 4.85ft = 18.85ft
W = 10ft + 4.85ft = 14.85ft
