The area of the rectangle below is represented by the expression 2x^2 - 3x - 20in^2. Determine the dimensions of the rectangle. Include units, and leave your answers as bionmials.

Question

I have no idea how to solve this, I don't even remember being taught this, so I can't really help you, sorry.....

Answer 1

This is much simpler than you think. All you need to realize is that the area of a rectangle A = L.w

where L is the length and w is the width. You are given the area A of the rectangle as a trinomial and in quadratic form as:

2x² -3x -20 in² . This trinomial is of the form ax²+bx+c and a product of two binomial terms ( ax + b)(cx + d)

with a = 2 , c =1 you get ( 2x + b)(x + d) now you have to find two numbers b and d whose product bd = -20, and the sum of the product of the outer terms is 2xd + bx = -3x. factor x outside x( 2d +b) = -3x divide by x both sides

2d + b = -3 , and bd = -20, and ask yourself what are two numbers I can multiply and get -20, the possible pairs ( b,d) are:

(5,-4), (-5,4), (10,-2), (-10,2), (1,-20),(-1,20) one of these pairs must give the solution

try (5,-4) , 2(-4)+5 = -3 therefore b = 5, and d = -4

Substituting you get ( 2x+5)(x-4) expand by foil method and you get 2x² - 4(2x) +5x -20 = 2x² -8x +5x -20 = 2x² -3x -20 and therefore the two binomials ( 2x + 5) ( x-4) are the correct binomial product = L.w

Now which one is L and which is w?

L > w as L is the length and w is the width. For any value of x, 2x+5 > x -4 and therefore

L = (2x+5) in and w = (x -4) in