## seattle12345 3 years ago Can someone please help with this?: The length of a rectangle is 2 inches longer than the width. If the area is 320 square inches, find the rectangle's dimensions. Round your answers to the nearest tenth of an inch.

1. amistre64

how do we determine that area of any rectangle?

2. msayer3

you have a problem with two unknown variables - x and y one being width and other being length - and two conditions 1) length being 2 inches longer than width and 2) area is 320.

3. seattle12345

@amistre64 Length x Width

4. amistre64

correct :) using that and the given information, we can determine the rest. Area = Length x Width ; Area = 320, and it says the Length is 2 times the Width 320 = (2xWidth)xWidth 320 = 2xW^2

5. amistre64

i read 2 times larger for some reason ....

6. amistre64

same concept tho; just a correct reading :) Area = Length x Width ; Area = 320, and it says the Length is (Width + 2) 320 = (Width+2)xWidth 320 = W^2 + 2W

7. seattle12345

would it be (W)(W+2)=320 ?

8. amistre64

yes

9. seattle12345

then would you subtract 320 from both sides?

10. amistre64

yes, and expand the left side; this turns it into a usual looking quadratic as a result

11. seattle12345

so it would be: W^2+2W-320=0?

12. amistre64

correct, and then you just use whatever method you like for the determining "w" complete the square, or quadratic formula

13. amistre64

since the "c" part is relatively overpowering; id opt complete the square if i had to do by hand w^2 + 2w +1 -1 -320 = 0 (w+1)^2 -321 = 0 (w+1)^2 = 321 w+1 = +- sqrt(321) w = -1 +- sqrt(321)

14. amistre64

then use a calculator to work out the decimal :) and ignore the negative since negative length is not defined

15. seattle12345

yea complete the square seems better in this situation. I always seem to miss a step with the quadratic formula. So W=16.91 and L=18.91?

16. amistre64

nearest tenth; but yes

17. seattle12345