rectanglegarden is 25ftX18ft. gravel path equal width around garden.

Question

How wide can the path be if there is enough gravel for 408ft^2?

We are using the quadratic equation I believe

Answer 1

draw a 25 by 18 rectangle

uniformly around that rectangle draw another rectangle

let x=the width of the gravel path

the dimensions of the larger rectangle are (18+x+x) by (25+x+x), or (18+2x) by (25+2x)

the area of the larger rectangle minus the area of the smaller rectangle equals the area of the gravel path

(2x+18)(2x+25)-(18)(25)=408

4x²+36x+50x+450-450=408

4x²+86x=480

4x²+86x-408=0

divide both sides by 2

2x²+43x-204=0

you can use the QE if you like but it's not necessary

2*204=2*2*2*3*17 and 3*17-8=51-8=43, the coefficient of the middle term

(2x   )(x   )=0 so far...

(2x   51)(x   4)=0 so far...

(2x+51)(x-4)=0

x-4=0

x=4 feet for the width of the gravel path

check: (18+8)(25+8)-(18)(25)=408

26*33-18*25=408

858-450=408

408=408