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Dejan
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Jan-08-02, 08:35 PM (EST) |
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"simple calculus or not?"
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Hi there, It was a long time ago since I have been doing calculations like this one and I forgot most of it. I am not sure does this question fit here, but here it is: Is it possible to find value for “X” from this?
26 = {X – (26 +X)/5}square + {10 – (26 +X)/5}square + {7 - (26 +X)/5}square + {4 - (26 +X)/5}square + {5 - (26 +X)/5}square
I know that "X" = 4 but could you please show all workings? Thanks
Dejan
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Laocon
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Jan-14-02, 07:54 AM (EST) |
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4. "RE: simple calculus or not?"
In response to message #2
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Further... Multiply out the brackets and group all similar terms (eg powers of x) together. Then you will need to find the order of the equation and use the best method to resolve the problem. EG Quadratic (order 2) can be best solved using inspection (by simplifying the equation into (x-a)(x-b) = 0) or the formula: x equals minus b plus or minus the square-root of (b squared minus 4 lots of a by c), all over twice a. You could also try graphing the function,after simplifying to F(x) = 0 to find when it crosses the x-axis... yadda yadda yadda.... |
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Dude
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Jan-13-02, 07:54 AM (EST) |
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3. "RE: simple calculus or not?"
In response to message #0
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Your original problem 26 = {X – (26 +X)/5}square + {10 – (26 +X)/5}square + {7 - (26 +X)/5}square + {4 - (26 +X)/5}square + {5 - (26 +X)/5}square can be rewritten as
26*25=(4*x-26)^2+(24-x)^2+(9-x)^2+(x+6)^2+(x+1)^2 The right hand side can be expanded and we find
that x must satisfy the equation
650=20*x^2-260*x+1370 Bringing the 650 to the other side and dividing by 20 gives us
x^2-13*x+36=0
factoring this we get
(x-4)*(x-9)=0 This shows us that the original problem has 2 solutions
x=4 and x=9.
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Gunes
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Jan-14-02, 03:00 PM (EST) |
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5. "RE: simple calculus or not?"
In response to message #0
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hmm, i am trying to understand what is supposed to be interesting about this problem. clearly you have a sort of distance equation in 5-dimensional euclidean space, and you have many 5's or relations to 5 (square(5)+1=26, square(5)-1=4). you also have that one of the points would be the diagonal of a 5-dimensional hypercube. but some of the numbers seem totally arbitrary in the '5' context' ...
perhaps, you are not merely trying to refresh your memory of highschol algebraic manipulations, but are rather trying to find a geometric short cut to this problem?  |
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