- Thread starter ptebwwong
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I believe you mode a error, but my math could be wrong. Please check my algebra

f[x]= 2/[x-4] + 3 / [x+1]

find x when f[x]=2

2= 2/[x-4] + 3/[x+1]

place over common denominator [x-4][x+1]

2= {2[x+1]+3[x-4] }/[x-4][x+1]

2[x-4][x+1]= 2x+2+3x-12

2[x^2-3x-4]= 5x-10

2x^2-6x-8=5x-10

2x^2-11x +2=0

x=11+/-[121-16]^1/2 all over 4

x=11+/- 10.25 all over 4

x= 21.25/4

x=.75/4

x=5.31 or x= .19

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arthur ohlsten said:... Please check my algebra ... There are some missing grouping symbols, Arthur; otherwise, it looks good.

2 = 2/[x - 4] + 3/[x + 1]

place over common denominator [x - 4][x + 1]

2 = {2[x + 1] + 3[x - 4]}/{[x - 4][x + 1]}

2[x - 4][x + 1] = 2x + 2 + 3x - 12

2[x^2 - 3x - 4] = 5x - 10

2x^2 - 6x - 8 = 5x - 10

2x^2 - 11x + 2 = 0

x = 11 +/- [121 - 16]^(1/2) all over 4

x = 11 +/- (105)^(1/2) all over 4

x = 11/4 + ?105/4

x = 11/4 - ?105/4

x?5.31 or x?.19

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Your algebra is correct

Arthur

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arthur ohlsten said:Your algebra is correct This is an unreferenced pronoun, Arthur. Are you talking to yourself?

Arthur

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mmm.....

I am afraid my degrees are in Physics, Math, and Engineering, not in English.

I appreciate your correction

Arthur