Question

The length of a rectangle is 7 inches more than its width. The area of the rectangle is equal to 2 inches less than 5 times the perimeter . What is the length and width

Answer 1

Let the width be w, then the length is w+7 units.

The area of the rectangle is $A=w(w+7)= w^{2}+7w$.

The perimeter of the rectangle is: 

P = 2(Width + Length)=2(w+w+7)=2(2w+7)=4w+14

"The area of the rectangle is equal to 2 inches less than 5 times the perimeter." means that:

A = 5P - 2

$w^{2}+7w=5(4w+14)-2$

$w^{2}+7w=20w+70-2$

$w^{2}-13w-68=0$

to solve the quadratic equation, let's use the quadratic formula

let a=1, b=-13, c=-68

$D= b^{2} -4ac=169-4(1)(-68)=169+272=441$

the root of the discriminant is 21

the roots are

w1=(13+21)/2=34/2=17
and
w2=(13-21)/2=-8/2=-4, which cannot be the width.

The width is 17 units, and the length is 17+7=24 units


Answer: w=17, l=24