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tregor
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Posted on 03-01-06 1:15
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can anyone solve: x^2 - 2^x =0 by algebraic method??? (x squared minus two to the power x equals zero) of course, without using graphing calculator.
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Eminem8
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Posted on 03-01-06 1:21
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x = 2 satisfies the equation ......
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nepalichhoro
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Posted on 03-01-06 1:29
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isnt that a basic alzebra , x = 2 and 4
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Hunk_in_Grave
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Posted on 03-01-06 1:35
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Take a logarithm of the equation, then segregate the x terms and the constants. Then take the derivative and solve the resulting equation for x. HTH
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nepalichhoro
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Posted on 03-01-06 1:47
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hunk ji kasari ho ?? x^2 = 2^ x 2logx = xlog2 2logx = 0.301x 2(d/dx)logx = 0.301 (d/dx) x 1/x = 0.1505 x= 6.64 milena ni ta , kasari ho yeso gardine dukha garne ho ki
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*
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Posted on 03-01-06 1:52
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x = 0 and 2 x^2-2x = 0 x^2-2x+1=1 (x-1)^2=1 X=2
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nepalichhoro
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Posted on 03-01-06 1:57
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* bro , teslai bhanchha "fluke" , process is wrong, answer is luckily right . MCQ bhaidiyeko bhaye ta thyakkai mileko ni yaar
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timetraveller
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Posted on 03-01-06 1:57
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hina haina, nepali chori, you found the defivative, i.e the maxima of the curve. it is a smooth curve which is easy to show as it has derivtives for all values of x. but 2, and 4 arents the only points.
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kundale
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Posted on 03-01-06 2:47
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for this particular curve the minima is the solution.......
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sujanks
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Posted on 03-01-06 4:43
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panpate
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Posted on 03-01-06 4:47
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sujanks ko herda ta malai aaile 4 class ko yad aayo ni maile tyo ta char class ma nai gareko hai¨¨ hum............
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KaLaNkIsThAn
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Posted on 03-01-06 4:51
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Sujan, thats Brilliant. ehe!!! Verify garnu paryo... :D
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nepesahila
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Posted on 03-01-06 4:56
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Panpate, You learned that in 4th grade? I think I did it in 11th. However, there is a mistake in that solution, I hope sujanks will finally let others know about it, or someone else can find it.
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sujanks
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Posted on 03-01-06 4:57
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4 hoina mitra.. 6 class ma hola..., first time algebra.. class 6 ma padauthyo kyara.. unless you went to school that started algebra in class 4
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panpate
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Posted on 03-01-06 5:01
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My informations stand as before. Tyo 4 mai ho tara ke garne aaile dimag budo bhayechha mistake nai patta lagauna sakina yar....
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nepalichhoro
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Posted on 03-01-06 5:55
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Lau kasle kahile padheko thaha chhaina tara let me explain where it is wrong : when you take the roots { Sq. Rt( 4-(9/2))^2} , there are + and - values on both side . after that its not you who take either positive or negatinve on both side . You take the sign on one side and its mathematics that guides what sign you have to take on other side. Well any way this did remind me of my school days and do keep on throwing these memories of past.
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sujanks
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Posted on 03-01-06 6:20
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yupe, the solution violates the law of indices. the square root of any positive number results in a positive and negative number of same value. so, the square root of 4 is not necessarily equal to 2, it could be -2. . .
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Pink
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Posted on 03-01-06 6:56
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Is that the question look like? then I solved it! :)
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DC_virus
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Posted on 03-01-06 7:41
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x^2-2x = 0 x^2-2x+1=1 (x-1)^2=1 x-1 = 1 OR x-1 = -1 therefore x =0 or x =2 yesari kina mildaina??? bujhena ali!!! dimag ali slow bhayo..mafi pau
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DC_virus
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Posted on 03-01-06 7:43
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aye sorry..question nai galat padhechu!! oops
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