Respuesta :
Let the width be w, then the length is w+7 units.
The area of the rectangle is [tex]A=w(w+7)= w^{2}+7w [/tex].
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
[tex]w^{2}+7w=5(4w+14)-2[/tex]
[tex]w^{2}+7w=20w+70-2[/tex]
[tex]w^{2}-13w-68=0[/tex]
to solve the quadratic equation, let's use the quadratic formula
let a=1, b=-13, c=-68
[tex]D= b^{2} -4ac=169-4(1)(-68)=169+272=441[/tex]
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
The area of the rectangle is [tex]A=w(w+7)= w^{2}+7w [/tex].
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
[tex]w^{2}+7w=5(4w+14)-2[/tex]
[tex]w^{2}+7w=20w+70-2[/tex]
[tex]w^{2}-13w-68=0[/tex]
to solve the quadratic equation, let's use the quadratic formula
let a=1, b=-13, c=-68
[tex]D= b^{2} -4ac=169-4(1)(-68)=169+272=441[/tex]
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
