SOLUTION: The length of a rectangle is 2 more than the width. The area is 24 square inches. Find the dimensions. It asks for the Length and width in inches as answers. I appreciate any hel

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 2 more than the width. The area is 24 square inches. Find the dimensions. It asks for the Length and width in inches as answers. I appreciate any hel      Log On


   

Question 442907: The length of a rectangle is 2 more than the width. The area is 24 square inches. Find the dimensions.
It asks for the Length and width in inches as answers. I appreciate any help with this, thank you.
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
Remember, the area of a rectangle is A+=+lw
They give us the area, A=24in%5E2.
We know that the length is 2 more than the width, so l+=+w%2B2
So 24+=+lw++=+%28w%2B2%29%2A%28w%29
24+=+w%5E2+%2B+2w
w%5E2+%2B2w+-+24+=+0
What multiplies to -24 and adds to 2? Since our product is negative and our sum is positive, then we must have a negative and positive answer.
Here are the factors of 24:
1 24
2 12
3 8
4 6
Notice 6 -4 = 2 and 6*-4 = -24.
Then we can write w%5E2+%2B2w+-24+=+%28w%2B6%29%28w-4%29
This gives us answers of w=-6w=4
But does it make sense to have a negative length? No.
So w=4 is the only LOGICAL answer.
Now plug w=4 back into either our area formula or l=w+2. I like adding. :P
l+=+4+%2B2++=+6
So our dimension are 6x4, or highlight%28l=6in%29 highlight%28w=4in%29.
To check:
6 is 2 greater than 4.
6in * 4in= 24 square inches.