The length and the width of the rectangle are 10 inches and 6 inches respectively.
In the question, we are given that the length of a rectangle exceeds its width by 4 inches, and the area is 60 square inches.
We are asked for the length and the width of the rectangle.
We assume the width (w) of the rectangle to be x inches.
Its length (l), exceeds its width (w) by 4 inches.
Thus, its length (l) = x + 4 inches.
Now, the area can be calculated using the formula, A = l*w, where A is its area, l is its length, and w is its width.
Thus, the area = (x + 4)x = x² + 4x.
But, we are given that the area is 60 square inches.
Putting the value, we get a quadratic equation:
x² + 4x = 60.
or, x² + 4x - 60 = 0,
or, x² + 10x - 6x - 60 = 0,
or, x(x + 10) - 6(x + 10) = 0,
or, (x - 6)(x + 10) = 0.
By the zero-product rule, we get:
Either, x - 6 = 0, or, x = 6,
or, x + 10 = 0, or, x = -10, but this is not possible as the width of a rectangle cannot be negative.
Thus, the width = x = 6 inches.
The length = x + 4 = 10 inches.
Thus, the length and the width of the rectangle are 10 inches and 6 inches respectively.
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