The lengthof a rectangle is 2 m more than its width. The area of the rectangle is 80 m^2. What is the length and the width of the rectangle?
Let the length be L. Now length is 2 more than width, so width is equal to L - 2.
Now we have the area as 80 m^2
Area = length * width = L ( L - 2) = 80
=> L^2 - 2L = 80
=> L^2 - 10L + 8L - 80 = 0
=> L(L - 10) + 8( L - 10) = 0
=> ( L + 8) ( L - 10 ) =0
So L can be 10 or -8.
But as length is not negative we eliminate -8.
Therefore the length is 10 m and width is 8 m.
We'll note L the length and W the width.
We'll write mathematically the first sentence:
L = 2 + W
The area of the rectangle is:
A = Length*Width
From enunciation, we know that the area is A = 80 m^2.
80 = (2 + W)*W
We'll remove the brackets:
80 = 2W + W^2
We'll use symmetric property:
W^2 + 2W - 80 = 0
We'll apply the quadratic formula:
W1 = [-2 + sqrt(4 + 320)]/2
W1 = (-2 + 18)/2
W1 = 8
W2 = (-2-18)/2
W2 = -10
Since it is about a length, we'll reject the negative value. We'll keep the positive value for W = 8 m.
The length is:
L = 2 + W
L = 2 + 8
L = 10 m
The length of the rectangle is 2 feet more than the width.
So we presume the width to be x and length x+2.
Therefore the area of the rectangle = x(x+2) which is actually given to be 80m^2.
=> x(x+2) = 80.
= x^2+2x-80 = 0.
=> (x+10)(x-8) = 0.
=> x+10 = 0. Or x-8 = 0.
=> x= -10. Or x = 8.
So width = x= 8 feet, length = x+2 = 8+2 = 10 feet.