Question

The area of a rectangle is 72 square inches. If the length is 40 inches greater than four times the width, what are the dimensions of the rectangle?

Answer 1

A=w*l
72=(40+4w)w
72=40w+4w2
4w2 +40w-72=0 divide by 4
w2 +10w -18=0
w_1,_2=(-10+/- sqrt(100+72))/2=(-10+/- sqrt172/2

Answer 2

but because one lenght of side not can be minus so than just (-10+sqrt172)/2 will be right

Answer 3

- so this dimension for width but now with this you can calcule easy the lenght too

Answer 4

so makeing this a quad equation and geting the -w im still on the right track??

Answer 5

The positive solution = 1.55743852" is the width of the rectangle
then
4(1.55743852) + 40 = 46.22975408" is the length
dose that look right or am i totaly missing the boat here

Answer 6

you can check it so if 72=(1,55)x(46) than those are right