Question

What is the length and width of a room that is 60 square meters and has a perimeter of 32 meters

Answer 1

L = lenght
W = width

$L * W = 60 \\ 2L + 2W=32\\\\ L = \frac{32-2W}{2}\\ \frac{32-2W}{2}*W = 60\\ (16-W)*W = 60\\ -W^{2} + 16W -60 = 0\\\\ d = 16^{2} - 4*1*60 = 256 - 240 = 16 \\\\ W1 = \frac{-16+\sqrt{16}}{-2} = \frac{-12}{-2} = 6\\ W2 = \frac{-16-\sqrt{16}}{-2} = \frac{-20}{-2} = 10\\ L1 = 10\\ L2 = 6$

so

{L,W} = {6,10} or {10,6}

but L > W  so

L = 10
W = 6