# The length of a rectangle is 6 inches longer than its width. what are the possible widths if the area

###### Question:

## Answers

Area=l×w

l=w+6

667=w(w+6)

w²+6w-667=0

w=23 or -29

w=23 only because distance cannot be negative

width must be atleast 23 inches.

Step-by-step explanation:

Given that the length of a rectangle is 6 inches longer than its width.

If l is length, then w = l-6

Area =lw

i.e. Area [tex]= l(l-6)\geq 667\\l^2-6l-667\geq 0\\(l-29)(l+23)\geq 0[/tex]

For this inequality being the product of two numbers is positive if both terms have the same sign.

This is possible only if l is atleast 29 inches (ignoring negative solution for l)

Hence possible widths are atleast 23 inches.

For a rectangle:A = Length x WidthL = W + 6A = ( W + 6 ) x WA = W² + 6 WW² + 6 W - 667 = 0W 1/2 = ( - 6 +/- √(6² - 4 · 1 · (-667)) / 2W 1/2 = ( - 6 +/- √ ( 36 + 2,668 ) ) / 2W = ( - 6 + √2,704 ) / 2 ( another solution is negative )W = ( - 6 + 52 ) / 2W = 46 / 2 W = 23 inches ( at least )If we want to prove it: W = 23, L = 23 + 6 = 2923 x 29 = 667

The possible widths are 23 inches and more.

1.1 and 18

2.factor x^2+4x-437

3.factor x^2-36x-667