A rectangle has an area of 345 square yards. It’s width is seven yards less than twice the length. Find the measures of the length and width of the rectangle.

Answer 1

Hi,

Here is how I would solve this problem.

Area = 345 sq. yd.; given

L = length; define variable for length

W = width; define variable for width

Area = L*W; formula for area of rectangle

W = 2*L-7; translate word problem into an equation

345 = L*(2L-7); substitute know items into formula for area of a rectangle

345 = 2L^2-7L; use distributive property

2L^2-7L-345 = 0; rearrange to get a polynomial equation in standard form

To find the factors I will find the factors of a*c = -690 that add up to b = -7

This is -30 and 23.

2L^2-30L + 23L-345 = 0; use these factors to rewrite the b term of the polynomial

2L(L-15) + 23(L-15) = 0; factor first two terms and last two terms (factor by grouping)

(2L+23)(L-15) = 0; use the distributive property to factor out (2L+23)

Use zero product property to find

L = 15

L = -23/2

Since L or length can only be positive we will choose 15

L = length = 15 yards

W = 2*L-7 = 30-7 = 23 yards

Regards,

Ray