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This thread was created on March 30, 2008
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Topic ID: 27955
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ana_is_a_banana
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 Posted: Sun Mar 30, 2008 5:26 pm Post subject: Algebra 1 |
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okay...
I had a problem:
The width of a rectangular park is 5m shorter than its length. If the area of the park is 300m². Find the length and the width.
I made this equation:
x(x+5)=300
And here are my Steps:
x(x+5)=300
x²+5x=300
X²+5x-300=0
x=-5±√5²-4(1)(300)/2
x=-5±√25-1200/2
x=-5±√-1175/2
okay what do I do from there? I thought a square root of something negative was impossible? Oh and the slashes mean it is a fraction. HELP PLEASE!!!!
These are all "solving problems involving quadratic equations" and I need major help with them!!!
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Ana
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Etude
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| Posted: Sun Mar 30, 2008 7:46 pm Post subject: |
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Feel free to enlighten us on how you got:
x=-5±√5²-4(1)(300)/2
x=-5±√25-1200/2
x=-5±√-1175/2
[pre:e1e156653f]x * ( x + 5 ) = 300 You've accomplished this
x^2 + 5x - 300 = 0 You've also accomplished this
( x + 20 ) * ( x - 15 ) = 0 The product of 20 and 15 will give you 300;
Once you FOIL it, you are given a positive
5x--everything works out, the process to find
that was just to find two numbers with a
difference of 5 that multiplied to 300
I find it to be the easiest way
x + 20 = 0 x - 15 = 0
- 20 - 20 + 15 + 15
x = -20 x = 15 The solutions(s). Unfortunately you cannot
have a negative distance, so -20 is out
of the question. Thus you have two dimensions:
15m x 20m.
Completion.[/pre:e1e156653f] |
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Leja
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| Posted: Sun Mar 30, 2008 9:06 pm Post subject: |
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| Whoops; I think I see where you made your error. If x represents the length of the rectangular park, then 5 should be subtracted, not added for the width to be a smaller value. I think that should clear up your problem with having a negative answer? |
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ana_is_a_banana
Junior Writer
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| Posted: Sun Mar 30, 2008 10:17 pm Post subject: |
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thanks!
I already figured it out, my friend helped me!!  |
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| This thread was created on March 30, 2008 |
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