Question

The length of a rectangle is 7 yd more than twice the width, and the area of the rectangle is 99 yd2 . find the dimensions of the rectangle.

Answer 1

The first thing we must do for this case, is to define a variable.
 We have then:
 w: width of the rectangle.
 By definition, the area of the rectangle is the product of the length by the width.
 We have then:
 $A = (w) * (2w + 7)$
 Substituting the value of the area we have:
 $99 = (w) * (2w + 7)$
 Rewriting the polynomial:
 $2w ^ 2 + 7w - 99 = 0$
 When factoring we have:
 $(w + 9) (2w-11) = 0$
 Taking the positive root we have:
 $2w-11 = 0 2w = 11 w = 11/2 w = 5.5 yd$
 Then, the length of the rectangle is:
 $2w + 7 = 2 (5.5) + 7 = 11 + 7 = 18 yd$
 Answer:
 the dimensions of the rectangle are:
 width = 5.5 yd
 Long = 18 yd