The top of a rectangular table has an area of 195 square feet. It has a length that is 2 feet more than the width. Find the dimensions of the table top.

Answer 1

Hello Monica,

I'd love to help you with this question.

So as we know an area of a rectangular table is going to be the product of the length times the width.

Let's represent that by length x width = area, or l x w = area for short.

In this question, the area of the table is 195 square feet, and the length is the width plus 2 feet.

So we can say:

l x w = 195 square feet

or replacing the length by width plus 2,

w(w+2) = 195

where:

w = width

w+2 = length

If we expand w(w+2) = 195, we get:

w^2 + 2w = 195 (w squared plus 2 times w equals 195)

It's a quadratic equation.

We just need to solve for w, knowing that whatever the width w is, the length is w+2

We can bring all the numbers of this quadratic equation to one side to help us solve it.

w^2 + 2w - 195 = 0

At this point we know that we'll have a better idea of what w equals once we factor the quadratic equation into a form such as (part a)x(part b) = 0, because either (part a) or (part b) will have to be null.

To do this, we need to find two numbers that multiply to (-195), and add to (+2), which is the coefficient of w in this equation.

The two such numbers that work are +15 and -13.

15 x (-13) = -195

15 + (-13) = 15-13 = 2

So now that we've found these numbers we can factor w^2 + 2w -195 = 0 to:

(w+15)(w-13)= 0

Meaning the product of (w+15) and (w-13) is 0.

If you expand this equation, because we have picked the right numbers, you will see that you get exactly w^2 +2w -195 = 0. :-) It is the exact same equation, just factored.

So now we know that in order for the factored equation to be true,

w+15=0

or

w-13=0

If w+15=0, w=(-15) feet. This is isn't possible because w is a width of a physical table, so it has to be positive.

If w-13= 0, w=13 feet. This makes sense.

Since the length l, is w+2, we have

width w = 13 feet

length l = 15 feet

13 feet x 15 feet = 195 square feet, so these numbers make sense.

We are done solving this question at this point, but there is a second way to do it.

I'm not sure if you've learnt this in school yet, but if you can't find the numbers you need to factor the quadratic equation, you can always solve it using the following general equation:

Where:

x is the unknown value(s) (in our case it was w for width, x=w)

a= coefficient of x^2 (in our case it was w^2 and the coefficient was 1)

b= coefficient of x (in our case x was w and the coefficient was 2)

c= the added number in ax^2 + bx + c =0 (in our case the added amount c was (-195))

In our case, we were just using w instead of x, for the width of the table.

In our above equation w^2 + 2w -195 =0,

a=1

b=2

c= -195

So we have 2 solutions for w, using the above equation.

The first solution is:

w = (-2 + sqrt(2^2- 4x1x(-195))) / 2x1

The second solution is

w' = (-2 - sqrt(2^2- 4x1x(-195))) / 2x1

Calculating this we get

w= (-2+28) / 2 = 26 / 2 = 13

w'= (-2-28) / 2 = (-30) / 2 = -15

We get the exact same solutions as above, when we used the factoring method!

In the same way, we know that w is the measurement of the width of a table, so it can be 13 feet, and it can't be -15 feet.

So we have the same answer as above, the width is 13 feet and the length is 2 feet more, 15 feet.

I really hope this helps. Let me know!