ATAR Notes: Forum
HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: Nagisa on December 09, 2012, 03:06:42 pm
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has roots
which are in geomoetric progression.
Show that
and hence show that 
can get first bit, but stomped on showing
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(x-a)(x-ca) &= x^3 - (ca + \frac{a}{c} - a)x^2 + (ca^2 + \frac{a^2}{c} - a^3)x - a^3<br />\\ x^3 + \frac{q}{p}x^2 } + \frac{r}{p}x + \frac{s}{p} &= x^3 - (ca + \frac{a}{c} - a)x^2 + (ca^2 + \frac{a^2}{c} - a^3)x - a^3<br />\\ \frac{s}{p} &= -a^3<br />\\ a &= \left(-\frac{s}{p}\right)^{\frac{1}{3}}<br />\\ <br />\\ \text{Since} \; x-a \; \text{is a factor of} \; P(x),} \; P(a) &=0<br />\\ P\left(\left(-\frac{s}{p}\right)^{\frac{1}{3}}\right) &=0<br />\\ \left(\left(-\frac{s}{p}}\right)^{\frac{1}{3}}\right)^3 + \frac{q}{p}\times\left(\left(-\frac{s}{p}\right)^{\frac{1}{3}}\right)^2 + \frac{r}{p}\times\left(-\frac{s}{p}\right)^\frac{1}{3} + \frac{s}{p} &=0<br />\\ -\frac{s}{p} + \frac{q}{p}\times\left(\frac{s}{p}\right)^{\frac{2}{3}} - \frac{r}{p}\times\left(\frac{s}{p}\right)^{\frac{1}{3}} + \frac{s}{p} &= 0<br />\\ \frac{q}{p}\times\left(\frac{s}{p}\right)^{\frac{2}{3}} &= \frac{r}{p}\times\left(\frac{s}{p}\right)^{\frac{1}{3}}<br />\\ \frac{q}{r} &= \left(\frac{s}{p}\right)^{\frac{1}{3}} \times \left(\frac{p}{s}\right)^{\frac{2}{3}}<br />\\ \frac{q}{r} &= \left(\frac{p}{s}\right)^{\frac{1}{3}}<br />\\ \frac{q^3}{r^3} &= \frac{p}{s}<br />\\ q^3s &= r^3p<br />\\ pr^3 - q^3s &=0 \end{aligned})
should be
, if you try the polynomial
, then,
whereas the other one gave -196560
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thanks man, didn't think of that, thought it would be fairly straight forward. btw, for everyone's sake. when using latex, could u pls use "\left(,\right)" instead of just "(,)". it's very impressive but just looks awful.
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yeah lol, I thought if I did \left( and \right), the lines would be too long, I shall fix that now
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cheers man, ill +1 you if you +1 me
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The respect system isn't meant to be as used a trade scheme. You +1 somebody if they do something worth reading on the forum.
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should be
, if you try the polynomial
, then,
whereas the other one gave -196560
yeah it's no mistake on my part, that's what the book says. dunno maybe its a mistake, can anyone explain? thanks again polar ya sickent