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| Math help in Algebra http://45.55.195.193/viewtopic.php?f=7&t=30851 |
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| Author: | ryry1237 [ Fri Apr 13, 2012 10:41 am ] |
| Post subject: | Math help in Algebra |
I've got this math problem in my head which I just can't seem to figure out. Question: "Find, in a + bi form, the two roots of the equation (1-x)^3 = x^3" I've already found one root... (1-x)^3 = x^3 (1-x) = x 1 = 2x x = 1/2 But how would I go about getting the other root? |
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| Author: | trystanr [ Fri Apr 13, 2012 3:07 pm ] |
| Post subject: | Re: Math help in Algebra |
(1-x)^3 = x^3 1-x^3=x^3 1=x^3+x^3 1=x^6 1=x If I'm remembering correctly |
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| Author: | Roast Veg [ Fri Apr 13, 2012 4:33 pm ] |
| Post subject: | Re: Math help in Algebra |
(1-x)^3 = 1-3x+3x^2-x^3 |
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| Author: | Fearful_Ferret [ Sat Apr 14, 2012 9:30 am ] |
| Post subject: | Re: Math help in Algebra |
trystanr wrote: (1-x)^3 = x^3 1-x^3=x^3 1=x^3+x^3 1=x^6 1=x If I'm remembering correctly u trollin? Anyway, Since the highest power of x is 3, there are 3 roots. As you found out, 1/2 is a root. Expanding out the (1-x)^3 term and bringing everything to the same side, we get 2x^3-3x^2+3x-1 = 0 Factoring out (x-1/2), we get (x-1/2)*(2x^2-2x+2)=0, or equivalently (x-1/2)*(x^2-x+1) = 0 So let's look at (x^2-x+1) = 0 Complete the square: (x^2-x+1/4)+3/4 = 0 So we get (x-1/2)^2+3/4 = 0 or (x-1/2)^2 = -3/4 Take the square root of both sides: x-1/2 = +-sqrt(-3/4) Leaving us with x = 1/2 +- i*sqrt(3)/2 QED (I kind of miss algebra now) |
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| Author: | Geti [ Sun Apr 15, 2012 6:18 am ] |
| Post subject: | Re: Math help in Algebra |
(take a course in it you'll stop missing it p quickly) |
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