candygirl200:

The length of a rectangle is 4 meters less than twice its width. The area of the rectangle is 70 m2 . Find the dimensions of the rectangle.

7 months ago
sammixboo:

Well, the formula to find the area of a rectangle is \(\color{red}{\sf wl = A}\), YOu were given a decent amount of information in the problem that we need to plug into the formula, such as the area, length, and width. The area is 70m\(\sf ^2\), while the length is 2w-4, and the width is w. Let's plug it in. \(\sf \color{red}{w(2w-4) = 70}\). Do you know what to do next?

7 months ago
candygirl200:

distribute the property and you would get 2w^2-4w=70

7 months ago
sammixboo:

Right, do you know the next step?

7 months ago
candygirl200:

move 70 to make it equal to 0

7 months ago
candygirl200:

2w^2-4w-70=0

7 months ago
candygirl200:

then u put it in a factor method which you will get 7 and -5 and u cannot have a negative distance so you make it a positive 5 and i think your answer will be 7 and 5 m is the dimensions right?

7 months ago
sammixboo:

You are definitely on the right track. You're right that we can't have a negative number for a distance, so therefor we will just toss -5 out the window. Our only option left, which is 7, will be our width. We still need to find the length. Do you know how we do that?

7 months ago
candygirl200:

yes make the 5 a positive

7 months ago
sammixboo:

No, we have to completely forget about the 5 since it is a negative number. We know that our width is 7, so now we have to find the length. The length is the for less than twice the width, so it's \(\sf \color{red}{2w-4=length}\). If we plug in 7 for w, then we get \(\sf \color{red}{2(7)-4=length}\), which is?

7 months ago
candygirl200:

10

7 months ago
sammixboo:

Right! So, our dimensions would be 7 for the width, and 10 for the length

7 months ago
candygirl200:

thanks!

7 months ago
sammixboo:

NO problem!

7 months ago