mcerd1
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| posted on 22/3/12 at 09:15 AM |
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OT - more maths help please
I really should be able to do this, but its been far too long since I had to do it and I was never that good at it so I keep making silly mistakes
what I've got is:
A = (2c + 2L)^2 - 4L^2
and then I start second guessing myself and ending up with 0=0 etc.....
i know that: (a+b)^2 = a^2 +2ab + b^2
but I forgotten how to deal with (2a + 2b)^2 [ = 4a^2 + 4b^2 + ??ab ]
cheers
-Robert
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wilkingj
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| posted on 22/3/12 at 09:41 AM |
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BODMAS is the Answer (Its what I was taught at school 50 years ago. It may not be modern, but its mathematically correct).
Do them in this order.
Brackets, Of, Division, Multiplication, Addition, Subtraction.
Hope that helps.
EDIT:
Hmm looks like a quadratic equasion.
There is a formula for them.
Also I think they are likely to have two answers for A and B.
But its been a long time!
[Edited on 22/3/2012 by wilkingj]
1. The point of a journey is not to arrive.
2. Never take life seriously. Nobody gets out alive anyway.
Best Regards
Geoff
http://www.v8viento.co.uk
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MikeRJ
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| posted on 22/3/12 at 12:11 PM |
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quote:
Originally posted by mcerd1
i know that: (a+b)^2 = a^2 +2ab + b^2
but I forgotten how to deal with (2a + 2b)^2 [ = 4a^2 + 4b^2 + ??ab ]
(2a + 2b)^2 = 4a^2 + 4b^2 + 8ab
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Macbeast
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| posted on 22/3/12 at 12:29 PM |
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Square the first, square the second, multiply together and double.
(2a + 2b ) ^2 = 4a^2 + 4b^2 + 8ab
( I think )
I'm addicted to brake fluid, but I can stop anytime.
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stevebubs
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| posted on 22/3/12 at 12:51 PM |
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quote:
Originally posted by mcerd1
I really should be able to do this, but its been far too long since I had to do it and I was never that good at it so I keep making silly mistakes
what I've got is:
A = (2c + 2L)^2 - 4L^2
and then I start second guessing myself and ending up with 0=0 etc.....
i know that: (a+b)^2 = a^2 +2ab + b^2
but I forgotten how to deal with (2a + 2b)^2 [ = 4a^2 + 4b^2 + ??ab ]
cheers
-Robert
2a x 2a = 4a^2
2a x 2b = 4ab
2b x 2a = 4ba = 4ab
2b x 2b = 4a^2
?? therefore = 8
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jonrotheray
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| posted on 22/3/12 at 02:17 PM |
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There's too many variables for this to be a quadratic.
but...
A=(2c + 2L)^2 - (2L)^2
Difference of squares ie a^2 - b^2 = (a-b)*(a+b)
A=(2c + 2L -2L)*(2c + 2L + 2L)
A=2c(2c + 4L)
A=4c^2 + 8cL
when A=0
4c^2 = -8cL
divide through by 4c
c = -2L
that's as far as you get (I think); there is no single set of values that satisfies the equation.
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mcerd1
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| posted on 22/3/12 at 03:37 PM |
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cheers guys - sorted again
quote:
Originally posted by jonrotheray
There's too many variables for this to be a quadratic.
c is the only real variable, the rest are constants for a given geometry / loading
and A can never be 0 for this case
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