The area of the opening of a rectangular window is to be 120 square feet. If the length is to be 2 feet more than the width, what are the dimensions

Width = 10ft

Length = 12ft

Step-by-step explanation:

For a rectangle of length L and width W, the area is calculated as:

A = L*W

In this case, we know that:

The area is 120 ft^2.

The length is 2 ft more than the width.

Then we can write:

L = W + 2ft.

If we use these two equations and replace them in the equation for the area of a rectangle, we get:

A = 120ft^2 = W*L = W*(W + 2ft)

120ft^2 = W*(W + 2ft)

Now we can solve this for W

120 ft^2 = W^2 + 2ft*W

Now we can rewrite this as:

W^2 + 2ft*W - 120ft^2 = 0

The solutions of this equation can be found by the Bhaskara's formula.

We know that for an equation of the form:

a*x^2 + b*x + c = 0

The two solutions are given by:

in our case, we have:

a = 1

b = 2ft

c = -120ft^2

Then the solutions for the equation W^2 + 2ft*W - 120ft^2 = 0 are:

Then the two solutions are:

W = (-2ft - 22ft)/2 = -12ft  (this is a negative measure, does not have a real meaning in this case, so this solution can be discarded)

The other solution is:

W = (-2ft + 22ft)/2 = 10ft  (this is the correct solution)

Then the width is 10ft, and we know that the length is 2 ft longer than the width, then the length is:

L  = W + 2ft = 10ft + 2ft = 12ft

answer: -2+√x,-2-√x

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