   SEARCH HOME Math Central Quandaries & Queries  Question from Nick, a student: A rectangle is twice as long as it is wide. If the length and width are both increased by 5cm, the resulting rectangle has an area of 50cm^2. Find the dimensions of the original. The equation I figured out was (2x + 5)(x + 5) = 50 FOIL: 2x^2 + 10x +25 = 50 2x^2 + 10x - 25 = 0 Now what? Hi Nick,

First of all (2x + 5)(x + 5) = 2x2 + 15x + 25 so your final equation is

2x2 + 15x - 25 = 0

At this point you can either use the general quadratic formula or you can complete the square. Let's complete the square.

2x2 + 15x = 2(x2 + 15/2 x) so to complete the square you need to add (15/4)2 = 225/16 inside the parentheses since x2 + 15/2 x + 225/16 = (x + 15/4)2. So here is my solution.

2x2 + 15x - 25 = 0
2(x2 + 15/2 x) - 25 = 0
2(x2 + 15/2 x + 225/16) - 25 - 225/8 = 0
2(x + 15/4)2 = 25 + 225/8 = 625/8
(x + 15/4)2 = 625/16
x + 15/4 = ±25/4
x = - 15/4 ±25/4

Thus x = 10/4 = 5/2 or x = -40/4. Since x is the length of a side of the rectangle it can't be negative so x = 5/2 cm.

I hope this helps,
Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.