math question solving trinomial equations?

Perimeter P= 64 cm
Area A=240 sq cm
2(l+w) = 64
l+w = 64/2 = 32
l = 32 -w

A = 240
lw = 240
substitute for l

(32 - w) w = 240
32w - w^2 = 240
w^2 - 32w + 240 = 0
w^2 - 12w -20w + 240 = 0
w(w - 12) - 20( w - 12) =0
(w-12)(w - 20) = 0
w =12 or 20
l =32 - w

for w = 12, l=32-12 = 20
for w=20, l=32-20 =12

so dimensions of the rectangle are 20 and 12

Length (x):
32 - x = 240/x
32x - x² = 240
x² - 16x = - 240 + (- 16)²
x² - 16x = - 240 + 256
(x - 16)² = 16
x - 16 = 4
x = 20

Width:
= 32 - 20
= 12

Answer: 20 cm & 12 cm are the dimensions.

Proof (area = 240 sq cm):
= 20 * 12
= 240

Proof (perimeter = 64):
= 2(20) + 2(12)
= 40 + 24
= 64

2l + 2w = 64 (solve by using substitution)
lw = 240

2l + 2w = 64
l + w = 32
l = 32 - w

lw = 240
(32 - w)w = 240
32w - w^2 = 240
w^2 - 32w + 240 = 0
w^2 - 12w - 20w + 240 = 0
(w^2 - 12w) - (20w - 240) = 0
w(w - 12) - 20(w - 12) = 0
(w - 12)(w - 20) = 0

w - 12 = 0
w = 12

w - 20 = 0
w = 20

lw = 240
l(12) = 240
l = 240/12
l = 20

lw = 240
l(20) = 240
l = 240/20
l = 12
(the length should be longer than the width)

∴ l (length) = 20 , w (width) = 12

The setting up isn't hard:

you got 2l+2w=w. getting the common factor out: 2(w+l)=64.
Then divide both sides by 2: w+l=32................ (1st equation)
From other equation: lw=240. divide both sides by l: l=240/w............(2nd equation)

Replace equation 2 on 1: w+(240/w)=32.
Get the common denominator w fro teh fraction: [(w^2+240)/w]=32.
Now multiply both sides by w: w^2+240=32w.
Set it equal to 0: w^2-32w+240=0.
Solve the quadratic equation. (I use the general form [(-b+sqrt.(b^2-4ac))/2a]). Or solve it how you better know how to do it.
and you get that w=20.
And finally replacing that on equation l=(240/w). you get that
l=12.
So the dimensions are w=20 and l=12
Hope it helps

The dimensions:

12x20cm

I used guess and check

You have the correct two equations. Solve the first for w in terms of l

2l + 2w = 64
2w = 64 - 2l
w = (64 - 2l)/2

Now, substitute this expression for w in the second equation.

l[(64 - 2l)/2] = 240

32l - l^2 = 240

solve the quadratic

l^2 - 32l + 240 = 0

(l - 12)(l - 20) = 0

l = 12 cm
then w = 20 cm

OR

l = 20 cm and w = 12 cm

Either set of solutions is correct.

QED

LW = 240
W = 240/L

2L + 2W = 64

By substitution,
2L + 2(240/L) = 64
L + 240/L = 32
L - 32 + 240/L = 0
L² - 32L + 240 = 0
(L-20)(L-12) = 0
L = 12, 20

The rectangle is 12 by 20 cm.

a+b=32
ab=240
X^2 -32X +240=0
X=16+4 & 16-4
a=20 ,b=12
==================================================
(x-20)(x-12)=x^2 - 32x +240