# Wierd quadratic world problem

#### hamza2095

##### New member
The problem: A rectangle is 60 x 80 (area is 4800), find what the width of a uniform border exactly half the area would be.

I came up with the equation 2400 = (80-2x)(60-2x) which then expands into 0 = 4x^2-280x+4800, I then did the quadratic equation which gave me the answer of 30 and 40 but both of these dont work because if it were 30 (6-2(30)) would equal 0 and if it were 40 (80-2(4)) would equal 0 and a dimension can't be 0. (this is how i came up with the equation)

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#### stapel

##### Super Moderator
Staff member
The problem: A rectangle is 60 x 80 (area is 4800), find what the width of a uniform border exactly half the area would be.

I came up with the equation 2400 = (80-2x)(60-2x) which then expands into 0 = 4x^2-280x+4800...
I agree with your setup and drawing (nice!). But how did you go from here:

. . . . .2400 = (80 - 2x)(60 - 2x) = 4800 - 160x - 120x + 4x^2

...to 0 = 4800 - 280x + 4x^2? #### hamza2096

##### New member
I agree with your setup and drawing (nice!). But how did you go from here:

. . . . .2400 = (80 - 2x)(60 - 2x) = 4800 - 160x - 120x + 4x^2

...to 0 = 2400 - 280x + 4x^2? My bad. I forgot to subtract the 2400 from 0, although i did when I got my answer of 30 and 40 with the quadratic equation. Does this mean it isn't possible or something?

#### stapel

##### Super Moderator
Staff member
My bad. I forgot to subtract the 2400 from 0, although i did when I got my answer of 30 and 40 with the quadratic equation.
What do you mean by "my answer of 30 and 40"? Shouldn't x have one value (being the consistent width around the edges)?

Does this mean it isn't possible or something?
Um... Does which mean what isn't possible? #### ksdhart

##### Full Member
Well, you're solving a quadratic, so the only way you could conclude that the problem isn't solvable is if you find there are no real roots of the quadratic. If, as you found, the answers you get don't check with the original equation, then you can conclude that you made a mistake somewhere I'd begin by noting that your proposed solutions of x = 30 and x = 40 are, in fact, the solutions to the following quadratic:

4800 - 280x + 4x2 = 0

4(x - 30)(x - 40) = 0

But, as we know, the above quadratic is not the one you're meant to work with. You're meant to work with this:

4800 - 280x + 4x2 = 2400

Subtracting 2400 from both sides (and rearranging terms) gives us:

4x2 - 280x + 2400 = 0

What do you get when you solve that quadratic?

#### Ishuda

##### Elite Member
I would have thought the border would be outside the original rectangle.