Question

A landscaper has enough cement to make a patio with an area of 150 Sq ft. The homeowner wants the length of the patio to be 6 ft longer than the width. What dimensions should be used for the patio? Round your answer to the nearest tenth of a foot.

Answer 1

$A=lw=150\\l=w+6\\(w+6)w=150\\w^2+6w=150\\w^2+6w-150=0\\w=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\w=\frac{-(6)+-\sqrt{(6)^2-4(1)(-150)}}{2(1)}\\w=\frac{-6+-\sqrt{36+600}}{2}\\w=\frac{-6+-\sqrt{636}}{2}\\w=\frac{-6+-2\sqrt{159}}{2}\\w=-3+-\sqrt{159}>0\\w=-3+\sqrt{159}\\\\l=(-3+\sqrt{159})+6\\l=3+\sqrt{159}$

Width = -3 + sqrt(159)
Length = 3 + sqrt(159)