ALGEBRA | Engineering Mathematics

Let \(x\) be the width of the strip. Thus the new length of the rectangle is \( 120 + x\) and the new width is \(80 + x\).

The area of the old plot: \(A_o = 120 (80) = 9600 yd^2.\)

The area of the new plot is twice the area of the old plot. Thus, \(A_n = 2A_o\)

\(A_n = (120 + x)(80 + x)\)

\(A_n = 9600 + 200x + 80x^2 \)

\(9600 + 200x + 80x^2 = 2 (9600)\)

\(x^2 +200x-900 =0 \)

\(x = \dfrac{(-200 + \sqrt{200^2 - 4(1)(-9600) })}{(2(1))}\)

\(x = 40 \; yd \) ANSWER