a parking lot has a rectangular area of 40,000yd^2. The length is 200 yd more than twice the width. What are the dimensions of the lot? So far I have width=x length=2x+200 so A=lw so 40,000=(2x+200)(x)

Question

a parking lot has a rectangular area of 40,000yd^2. The length is 200 yd more than twice the width. What are the dimensions of the lot? So far I have width=x length=2x+200 so A=lw   so 40,000=(2x+200)(x)

anonymous · Answer

then i took 2x^2 +200x=40,000 everything after that is a mess

anonymous · Answer

answer should be 100x400 how do i get there?

anonymous · Answer

Set it equal to 0:
$$2x^{2} + 200x - 40000 = 0$$
Then factor:
$$(2x - 200)(x+200)$$

anonymous · Answer

$$x=100$$$$(x+200)(x-100)=0$$$$x^2+100x-20000=0$$