i am desperate can any one help?

Question

a rectangular table top has an area of 60 sq feet if the length of the table is one foot less than 4 times the width, find its dementions

Answer 1

Length times width equals area - L X W = A

A = 60

L = 4W - 1

Solve (I'm not going to do ALL your homework for you...)

Answer 2

Draw the rectangle on your paper. Ok, the area is 60 sq. feet, so

A=60 sq. ft. (remember that square feet = "feet squared" or "feet times feet")

Now, you need to create a couple of equations to help you.

On your paper, call the longer side of the rectangle "L" for length. And call the shorter side "w" for width.

Write down what you know. The problem states that the Length is one foot less than 4 times the width. So the width = w.

One foot less than 4 times the width in equation form is:

4w - 1

(4 times the width = 4w and less one foot give you the "-1" part).

So

L=4w-1

We also know that the area (A) of something is its Length times its width, so:

A= L x w

We also know that A=60, so

60 = L x w

We have an equation for L, so let's use substitution and put it in place of the "L" in the equation above:

60 = (4w - 1) x w

Let's replace the "x" (times) symbol with * (asterix) which is what we use ont he computer so we don't get "times" confused with "x" (the letter):

60 = (4w - 1) * w

Now multiply out the right side of the equation:

60 = 4w^2 - 1w
60 = 4w^2 - w

(w^2 is w squared or raised to the second power - which we call "squared")

Now you can use the quadratic equation to figure out "w":

(From the link below)

*IF* an equation is in the form A*x^2+B*x+C = 0, then the 2 solutions
for that equation (also called zeros because the right side is zero)
are given by the formulas (HAVING TROUBLE WITH THESE ON THIS PAGE - PLEASE SEE THE LINK IN THE SOURCES BOX BELOW MY ANSWER TO SEE HOW THEY ARE SUPPOSED TO LOOK!):

-B + sqrt(B^2 -4*A*C)
x = -----------------------
2*A

OR

-B - sqrt(B^2 -4*A*C)
x = --------------------------------
2*A

"sqrt" means "the square root of"

So subtract 60 from both sides of your equation:

60-60 = 4w^2 - w - 60

0= 4w^2 - w - 60

I'm going to turn the equation around for simplicity. Remember that Z=Y is the same thing as Y=Z (I'm just turning it around - no math involved here):

4w^2 - w - 60 = 0

Now, you can see that it's in the form we need:

(a)x^2+(b)x+(c) = 0

So a is 4, b is -1 and c is -60. Plug these into the equations (remember, you've got to do BOTH equations):


-B + sqrt(B^2 -4*A*C)
x = -----------------------
2*A

OR

-B - sqrt(B^2 -4*A*C)
x = --------------------------------
2*A

-(-1) + sqrt((-1)^2 -4*(4)*(-60))
w = -----------------------
2*(4)

OR

-(-1) - sqrt((-1)^2 -4*(4)*(-60))
w=--------------------------------------
2*(4)

Time to do the math. :)

(-1)^2 is just 1

4*4*-60 = -960 (watch your signs)

So under the sqrt in both equations, we have "sqrt 961" and if you do this on your calculator, you get 31 (yay! a nice even number! :)

So now the equations look like this:

-(-1) + 31
w = -----------------------
2*(4)

OR

-(-1) - 31
w=-----------
2*(4)



-(-1) = 1
2 * 4 = 8

So now we have:

1 + 31
w = -----------------------
8

OR

1 - 31
w=----------
8

More calculations:

32
w = -----------------------
8

OR

-30
w=----------
8
We can see that the equation on the bottom is going to give us a negative number as an aswer - this is a big red flag! We can't have a negative width!!! The width is there and we can measure it, so (in your universe) it's a positive number only. :) So disregard the right equation from this point on. It's not an answer to our problem. :)

Do the calculation on the left equation and we get

w = 32/8 = 4

Now, go waaayyyyy back to the beginning and remember that we still have to find L. You can plug w into either of the equations we found in the very beginning, but I think the easiest would be A = L x w:

A = L*w
60 = L * 4

Solve for L:

L= 60/4

L = 15

The dimensions of L=15 and W=4 would give us a rectangular table top. But we need to confirm that this is indeed our answer. So let's plug them into our OTHER equation from the beginning as a check:

L=4w-1

L=4(4)-1

L=16-1

L=15

Yep! It worked! :) So, you can list the dimensions of the table as

L=15 feet

and

w = 4 feet

:)

Answer 3

Let the length of the table be L ft and width be W ft.

Area of table top = LW = 60 sq ft ---- (1)

Also, they say the length of the table is 1 foot less than 4 times the width, this means,

L = 4W - 1 ---- (2)

Sub (2) into (1): (4W-1)(W) = 60
4W^2 - W - 60 = 0
Factorise the quadratic equation to get:
(4W + 15)(W - 4) = 0
W = 4 or -15/4 (rejected)
L = 15

Answer 4

length = 15, width = 4

Answer 5

call the width x
call the length 4x-1

x(4x-1)=60
4x^2-x-60=0
(4x+15)(x-4) = 0

x=4 (disregard the negative solution for x)

so width is 4
length is 4*4-1= 15

Answer 6

L = 4w - 1, area = Lw = (4w-1)w = 60, so
4w² - w = 60
4w² - w - 60 = 0
4w² - 16w + 15w - 60 = 0
4w(w - 4) + 15(w - 4) = 0
(4w + 15)(w - 4) = 0
so w = 4, L = 15

Answer 7

let the width=x feet
length=4x-1
area=x(4x-1)=60
4x^2-x-60=0
4x^2-16x+15x-60=0
4x(x-4)+15(x-4)=0
(x-4)(4x+15)=0
x=4,-15
neglecting the negative answer, we gget the answer:
width=4feet. ANS.
length=15ft.

Answer 8

x (4x-1) = 60, solve HINT- they are both whole numbers