Question

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.

Answer 1

how do we determine that area of any rectangle?

Answer 2

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.

Answer 3

@amistre64 Length x Width

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yes

Answer 9

then would you subtract 320 from both sides?

Answer 10

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

Answer 11

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

Answer 12

correct, and then you just use whatever method you like for the determining "w"

complete the square, or quadratic formula

Answer 13

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)

Answer 14

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

Answer 15

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?

Answer 16

nearest tenth; but yes

Answer 17

THANKS FOR YOUR HELP!!!!