quadratic eqns: rectangle's length 1 cm longer than width...

[jburgswife30]

The length of a rectangle is 1 cm longer than its width. If the daigonal of the rectangle is 4cm, what are the dimensions (length and width) of the rectangle.

L^2+W^2=D^2
(w+1)^2+W^2=(4)^2
(w+1)(w+1)+w^2=16
w^2+1w+1+w^2=16
2w^2+1w+1=16
2w^2+1w=15
2w^2+1w-16=0
(w-16)(w+1)=0
w-16=0
w=16
or w+1=0
w=-1

width =16cm
length= 17cm


Does this look right?

 

[rahidz2003]

Re: quadratic equations

The length of a rectangle is 1 cm longer than its width. If the daigonal of the rectangle is 4cm, what are the dimensions (length and width) of the rectangle.

L^2+W^2=D^2
(w+1)^2+W^2=(4)^2
(w+1)(w+1)+w^2=16
w^2+1w+1+w^2=16
2w^2+1w+1=16
2w^2+1w=15
2w^2+1w-16=0
(w-16)(w+1)=0
w-16=0
w=16
or w+1=0
w=-1

width =16cm
length= 17cm


Does this look right?

Can't be right, if the width is 16 and height is 17, the diagonal isn't gonna be 4

(w+1)(w+1)+w^2=16
becomes
w^2+2w+1+w^2=16, not w^2+1w+1+w^2=16

So 2w^2 + 2w + 1= 16
2w^2 + 2w = 15
2w^2 + 2w - 15 = 0

Use the quadratic formula to get

w = (-2 +/- sqrt(124))/4

One of the answers gets kicked out because it's negative, so w =

(-2 + sqrt(124))/4

which is around 2.284

Length = 1 + 2.284 = 3.284
Width = 2.284
Diagonal = 4

 

[Deleted member 4993]

Re: quadratic equations

The length of a rectangle is 1 cm longer than its width. If the daigonal of the rectangle is 4cm, what are the dimensions (length and width) of the rectangle.

L^2+W^2=D^2
(w+1)^2+W^2=(4)^2
(w+1)(w+1)+w^2=16
w^2+1w+1+w^2=16 <--- that should be w^2 + 2w + 1 + w^2 = 16
2w^2+1w+1=16
2w^2+1w=15
2w^2+1w-16=0 <---does not follow from above
(w-16)(w+1)=0 <--- does not follow from above
w-16=0
w=16
or w+1=0
w=-1

width =16cm
length= 17cm

Does this look right? - no - a rectangle with sides 16 and 17 will have diagonal larger than 17