Math Help - Quadratic equation help

Hi,

I have come up against a question in a past paper I am revising for a maths exam next week and I am having some difficulty.

The question is:

A concrete path with uniform width x metres is to be built around the rectangular base of a house of length 61 metres and width 35 metres. The area of the path is 400m^2. Set up a quadratic equation in x and solve it to give the width of the path.

So far I have:

(61+2x)(35+2x)-(61*35)=400

(2135+122x+70x+4x^2)-(2135)=400

192x+4x^2=400

192x+4x^2-400=0

48x+x^2-100=0

Originally Posted by cjmcdougall

Hi,

I have come up against a question in a past paper I am revising for a maths exam next week and I am having some difficulty.

The question is:

A concrete path with uniform width x metres is to be built around the rectangular base of a house of length 61 metres and width 35 metres. The area of the path is 400m^2. Set up a quadratic equation in x and solve it to give the width of the path.

So far I have:

(61+2x)(35+2x)-(61*35)=400

(2135+122x+70x+4x^2)-(2135)=400

192x+4x^2=400

192x+4x^2-400=0

48x+x^2-100=0

=> (x + 50)(x - 2) = 0 ....

Hello, cjmcdougall

A concrete path with uniform width metres is to be built around
the rectangular base of a house of length 61 m and width 35 m.
The area of the path is 400 m^2.
Set up a quadratic equation in and solve it to give the width of the path.

So far I have:

. . (61 + 2x)(35 + 2x) - (61)(35) .= .400

. . 2135 + 122x + 70x + 4x^2 - 2135 .= .400

. . 2135 + 122x + 70x + 4x^2 - 2135 .= .400

. . 192x + 4x^2 .= .400

. . 192x + 4x^2 - 400 .= .0

. . 48x + x^2 - 100 .= .0

And you don't know how to solve a quadratic equation?
. . Then why was this problem assigned to you?


You have: .x^2 + 48x - 100 = 0

Factor: .(x - 2)(x + 50) = 0

And we have: .x = 2, x = -50


The only realistic answer is: .x = 2 meters.

I was assigned this as on my course I am expected to already know this even though I have never been taught it.

Originally Posted by cjmcdougall

I was assigned this as on my course I am expected to already know this even though I have never been taught it.

In which case, you could consider another way
to lead you into solving quadratics....

Draw a diagram of the path surrounding the house
Continue all horizontal and vertical lines from the inner rectangle to the outer rectangle.

x^2+61x+x^2+35x+x^2+61x+x^2+35x = 400
2x^2+61x+35x = 200

2x^2+96x = 200

x^2+48x = 100

x(x+48) = 100

one of the factors is 48 more than the other.
Find the factors of 100 that differ by 48.

100 = 100(1)
100 = 50(2)

etc.

Originally Posted by cjmcdougall

I was assigned this as on my course I am expected to already know this even though I have never been taught it.

What would your teacher say to that?

Originally Posted by Wilmer

What would your teacher say to that?

I'm coaching two-thirds of the students from a class,
whose teacher has told them he will not be covering
a quarter of their entire syllabus.
The course is 2 years in duration.
It's quite common for teachers not to cover the syllabus in time.

Then you should talk to the chair of the department about that. The printed syllabus for a class, or description in the school catalog, is a legal contract. While a certain amount of "contraction" is acceptable, any school administrator, including chair of the department, should be very sensitive to the possiblility of a law suit. If a teacher covers only 3/4 of the syllabus, repeatedly, that is a very dangerous situation.