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June 25, 2025, 02:41:04 pm

Author Topic: cherylim23's methods question thread :)  (Read 15602 times)  Share 

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Re: cherylim23's methods question thread :)
« Reply #120 on: January 27, 2011, 11:17:41 am »
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How would you solve this question?



Apparently, taking log doesn't work, but by inspection, I could tell x = 0 or 1
(Inspection is dodgy method IMO!)
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Re: cherylim23's methods question thread :)
« Reply #121 on: January 27, 2011, 11:47:16 am »
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How would you solve this question?



Apparently, taking log doesn't work, but by inspection, I could tell x = 0 or 1
(Inspection is dodgy method IMO!)

I have no idea how to solve this either...

I did it graphically by graphing (addition/subtraction of ordinate method) and the x-ints were 0 and 1 (your solutions)... But algebraically, I have no idea. I am interested in any algebriac solution too!
« Last Edit: January 27, 2011, 12:21:39 pm by Rohitpi »

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Re: cherylim23's methods question thread :)
« Reply #122 on: January 27, 2011, 04:55:07 pm »
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I'm pretty sure it's a transcendental function. There are ways to solve it, namely Newton's method, but it's pretty complicated and can only get you approximations (although good approximations).
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Re: cherylim23's methods question thread :)
« Reply #123 on: January 27, 2011, 05:52:08 pm »
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How about rooting both sides by x? though the answer was only 1 ):
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Re: cherylim23's methods question thread :)
« Reply #124 on: January 28, 2011, 08:15:40 pm »
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yeah, don't think there is an easy solution:

http://www.wolframalpha.com/input/?i=2^x+%3D+x%2B1
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Re: cherylim23's methods question thread :)
« Reply #125 on: February 20, 2011, 08:18:22 pm »
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Just a confirmation,

Dilated by a factor of n value from x-axis = Dilated by a factor of n value parallel to y-axis = Dilated by a factor of n value in the y-direction = Dilated by a factor of n value about the x-axis?

Hope that didn't seem too confusing!
« Last Edit: February 21, 2011, 07:48:11 pm by cherylim23 »
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Re: cherylim23's methods question thread :)
« Reply #126 on: February 20, 2011, 08:23:36 pm »
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This 'Dilated by a factor of n value from x-axis = Dilated by a factor of n value parallel to y-axis' is true, i havent seen the other 2 tho. Also my teacher said that we dont need to know dilated by a factor of n parallel to x/y axis.

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Re: cherylim23's methods question thread :)
« Reply #127 on: February 20, 2011, 08:41:04 pm »
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Yes, they are equal. Though I should point out that the first and last one are the same o_O
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Re: cherylim23's methods question thread :)
« Reply #128 on: February 21, 2011, 07:45:49 pm »
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Yes, they are equal. Though I should point out that the first and last one are the same o_O

oh, sorry! Thanks for pointing out, I meant to say, is "Dilated by a factor of n from x-axis = Dilated by a factor of n about x-axis"?  :uglystupid2:
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Re: cherylim23's methods question thread :)
« Reply #129 on: February 21, 2011, 08:02:57 pm »
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This 'Dilated by a factor of n value from x-axis = Dilated by a factor of n value parallel to y-axis' is true, i havent seen the other 2 tho. Also my teacher said that we dont need to know dilated by a factor of n parallel to x/y axis.

In my opinion I think parallel to y axis is worse because it doesn't specify where the dilation is from (though in most cases it is from the x-axis), for example if I said what do you get if you dilate the graph from of by a factor of 2 from the line then the dilation is parralel to the y-axis but it is not from the x-axis. :P
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Re: cherylim23's methods question thread :)
« Reply #130 on: May 29, 2011, 05:22:33 pm »
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Stuck on a question!

Given that the expression leaves the same remainder whether divided by x-b or x-c where b =/= c, show that b + c = 5
Given further that 4bc = 21 and b > c, find the values of b and c.
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Re: cherylim23's methods question thread :)
« Reply #131 on: May 29, 2011, 05:56:11 pm »
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Ok so the same remainder is left implies that b2-5b+7=c2-5c+7 where c does not equal b
so b2-c2=5(b-c)
factorise LHS (b-c)*(b+c)=5(b-c)
now cancel the b-c
b+c=5

next part
4bc=21, b>c
so b=5-c
4(5-c)c=21
expand it 20c-4c2-21=0
so 4c2-20c+21=0
now factorise (2c-3)*(2c-7)=0
so c = 3/2 or c=7/2
therefore b= 7/2 or 3/2 but b>c
so b=7/2 and c =3/2
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