Quadratic equations

Marcus Clayson

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How do I solve this problem?
Two rectangular rooms each have an area of 240 square metres. If the length of one of the rooms is x metres and the other room is four metres longer, write down the width of each room in terms of x. If the widths of the rooms differ by 3 metres, form an equation in x and show hat this reduces to x^2+4x-320=0. Solve this equation and hence find the difference between the perimeters of the rooms.
 
How do I solve this problem?
Two rectangular rooms each have an area of 240 square metres. If the length of one of the rooms is x metres and the other room is four metres longer, write down the width of each room in terms of x. If the widths of the rooms differ by 3 metres, form an equation in x and show hat this reduces to x^2+4x-320=0. Solve this equation and hence find the difference between the perimeters of the rooms.
Did you follow the steps indicated in the problem?

  1. If the length of one of the rooms is x metres and the other room is four metres longer, write down the width of each room in terms of x.

  2. If the widths of the rooms differ by 3 metres, form an equation in x and show hat this reduces to x^2+4x-320=0.

  3. Solve this equation and hence find the difference between the perimeters of the rooms.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.
 
Dear Subhotosh

Many thanks for your prompt reply. I’m sorry I didn’t “follow protocol” in my original message as this was the first time that I had used this system. I hope that this post is better.

What I did was first draw two rectangles, A and B. I made the length of rectangle A x metres and the width x+3 metres. Rectangle B has a length of x+4 metres and a width of x metres.
This means that x(x+3) = 240 m2
and x(x+4) = 240 m2.
Therefore x2 + 3x +x2+4x = 480m2 – i.e. 2x2+7x-480=0

This cannot be factorised, so using the quadratic formula with a = 2, b=7 and c=-480, I got a positive x value of 13.84m. Substituting this into rectangle A, 13.84 x 16.84 = 233.07m2 which is not 240m2 as in the question.

For rectangle B I got 13.84 x 17.84m2 which equals 246.91m2

My figures are close but not correct. Can you tell what I have done wrong?
 
For the first room we have:

[MATH]xw_1=240\implies w_1=\frac{240}{x}[/MATH]
And for the second we have:

[MATH](x+4)w_2=240\implies w_2=\frac{240}{x+4}[/MATH]
We know we must have [MATH]w_2<w_1[/MATH] and so we may state:

[MATH]w_1-w_2=3[/MATH]
Or:

[MATH]\frac{240}{x}-\frac{240}{x+4}=3[/MATH]
Multiply by \(x(x+4)\) to get:

[MATH]240(x+4)-240x=3x(x+4)[/MATH]
This reduces to (distributing and arranging in standard form, then dividing out any common numeric factors):

[MATH]x^2+4x-320=0\quad\checkmark[/MATH]
To factor, we need to factors of -320 whose sum is 4, and these are 20 and -16, hence:

[MATH](x+20)(x-16)=0[/MATH]
Discarding the negative root (since \(x\) represents a physical distance) we conclude:

[MATH]x=16[/MATH]
This implies:

[MATH]w_1=15[/MATH]
[MATH]w_2=12[/MATH]
And so the perimeter of room 1 is:

[MATH]P_1=2(x+w_1)[/MATH]
[MATH]P_2=2((x+4)+w_2)[/MATH]
[MATH]P_2-P_1=2((x+4)+w_2)-2(x+w_1)=2(x+4+w_2-x-w_1)=2(4-(w_1-w_2))=2(4-3)=2[/MATH]
We see we didn't even need to know \(x\) to find the difference in perimeters, we just needed to know the difference in widths, which we were already given. :)
 
[MATH]x = \text {length of room 1}.[/MATH]
[MATH]x + 4 = \text {length of room 2}.[/MATH]
[MATH]\text {width of room 1} = \dfrac{240}{x}.[/MATH]
[MATH]\text {width of room 2} = \dfrac{240}{x + 4}.[/MATH]
[MATH]\dfrac{240}{x} > \dfrac{240}{x + 4} \ \because x > 0.[/MATH]
[MATH]\therefore \dfrac{240}{x} - \dfrac{240}{x + 4} = 3 \implies[/MATH]
[MATH]\dfrac{80}{x} - \dfrac{80}{x + 4} = 1 \implies[/MATH]
[MATH]80 - \dfrac{80x}{x + 4} = x \implies[/MATH]
[MATH]80(x + 4) - 80x = x(x + 4) \implies[/MATH]
[MATH]80x + 320 - 80x = x^2 + 4x \implies[/MATH]
[MATH]x^2 + 4x - 320 = 0.[/MATH]
EDIT Seeing what Mark has said, if you solve the quadratic above, which does factor, you get

[MATH](x - 16)(x + 20) = 0 \implies x = 16 \text { or } x = - 20, \text { which is nonsense.}[/MATH]
[MATH]\therefore \text { dimensions of room 1 are } 16 \text { by } 15, \text { and }[/MATH]
[MATH]\text { dimensions of room 2 are } 20 \text { by } 12.[/MATH]
To my mind, it is a whole lot easier to identify four unknowns and four numeric relationships and solve from there.
 
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This means that x(x+3) = 240 m2
and x(x+4) = 240 m2.
Don't you see that those two equations are contradictory?

1) x(x+3) =240
2) x(x+4) = 240

Note that x=x and x+3 does not equal x+4

So how could x(x+3) possible equal x(x+4)

Alternatively you could have seen the contradiction this way.
x(x+4) = x(x+3 + 1) = x(x+3) +x.

How could x(x+3) and x(x+3) + x both equal 240?
 
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