New quadratics - dimensions question

Question

By adding the same amount to its length and width, a contractor increased the area of a rectangular room by 165ft^2 to make it 17ft by 21ft. What were the original dimensions of the room?

I'm having trouble understanding on how to set up the equation. I can solve quadratics pretty easily, but I've always had trouble setting them up. Any help would be appreciated!

Answer 1

17*21 = 357

357 - 165 = 192 is the original area

Lw = 192

(l+x)(w+x) = 357

Lw + lx + wx + x^2 = 357

192 + Lx + wx + x^2 = 357

Lx + wx + x^2 = 165

17 = w + x --> 17 - x = w

21 = L+x ---> 21 - x = L

(21 - x)x + (17-x)x + x^2 = 165

21x - x^2 + 17x - x^2 = 165

38x - 2x^2 - 165 = 0

2x^2 - 38x + 165 = 0

( 2x - 33)(x - 5 ) = 0

2x - 33 = 0 or x-5 = 0

x = 33/2=16.5 or x = 5

If the new rectangle has dimensions 17 x 21 then the

original triangle has/had dimensions:

0.5 x 4.5 or 12 x 16

The former does not have area 192

but the latter does

The original dimensions are 12 x 16