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The width of a rectangle is 12 feet less than the length. The area of the rectangle is 540 square feet. Find the dimensions of the rectangle.

Length $=30$ feet;Width $=18$ feet

Algebra

Chapter 1

THE INTEGERS

Section 7

Quadratic Equations with Integral Roots

The Integers

Equations and Inequalities

Polynomials

Missouri State University

McMaster University

Harvey Mudd College

Lectures

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in this problem we have the width of a rectangle is 12 feet less than this length. So let's call this the with. We'll call this the length and we'll say that w equals L minus 12 were also give that the area is 540. So we know the area of a rectangle just the with times length we have w l equals 5 40 Now we can just substitute in w so that we have this equation in terms about doing that We're going to get l A. Times l minus 12 equals 5 40 Multiplying this out, we'll get l squared minus 12 l If you subtract this 5 40 minus 5 40 equals zero. So let's see, what can we factor this 5 40 into? We know that doing some simple calculations that 5 40 is actually a multiple of thirties. We can say that l minus or my bad plus 30. And then if we divide 5 40 by 30 it'll be 54 over three. So 18. So if we want to get a negative 12 l will actually want to have a negative 30 here. So, huh, Negative 30 and then l plus 18. So this is a bit of a trick you're one to find. But if you use the calculator is actually really easy. So it be 13 18 checking really quickly. Have l squared plus 18 on minus 30 out, which will give us negative about and then our minus 5 40 So, of course l cannot equal negative 18. That's a negative length toe. All must be 30. So knowing that l is 30 you can put that back into the with so 30 minus two of his 18 and you're with his 18.

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