Question

length of a rectangular yard is 4 ft longer than its width w. The area of the yard is 525 ft2. (a) Write a quadratic equation in terms of w that represents the situation. (1 point) (b) What are the dimensions of the yard? Show your work. (2 points)

Answer 1

The answer is
(a) w² + 4w - 525 = 0
(b) length = 25 ft; width = 51 ft

length of a rectangular yard is 4 ft longer than its width w:
l = w + 4
 
The area of the yard is 525 ft2: A = 525 = l * w

(a) So:
l * w = 525
(w + 4) * w = 525
w² + 4w = 525
w² + 4w - 525 = 0

(b) Solve quadratic function:
w = (-b +/-√(b² - 4ac)/(2a)
    = (-4 +/-√(4² - 4 * 1 * (-525)))/(2*1)
    = (-4 +/-√(16 + 2100))/2
    = (-4 +/-√2116))/2
    = (-4 +/- 46)/2

w = (-4 - 46)/2 = -50/2 = -25       width cannot be negative, so w ≠ -25
or
w = (-4 + 46)/2 = 42/2 = 21

Since l = w + 4 and w = 21, then l = 21 + 4 = 25.

Answer 2

Answer:

Length 25ft, Width 21 feet

Step-by-step explanation:

l=w+4 (in terms of w: (w+4)*w=525)

A=l*w

l*w=525ft sq

now you have a quadratic, choose AC method X^2+bx+c

w^2+4w=525 (subtract 525 from both sides to make = to 0 to fit formula)

w^2+4w-525=0  (find a pair of numbers whos product =-525 and sum is 4)

-21,25

(w-21)(w+25)=0

W-21=0    or   W+25=0, so W= 21 or -25  you can not have negative length or width so choose 21 (width =21)

l=25