Author Topic: Halp pls (Read 2800 times) Tweet Share

Nagisa · « **on:** December 09, 2012, 03:06:42 pm »

$P(x)=px^{3}+qx^{2}+rx+s=0$ has roots $ac, a, \frac{a}{c}$ which are in geomoetric progression.

Show that $a=3\sqrt\left(-\frac{s}{p}\right)$ and hence show that $pr^{3}-qs^{3}=0$

can get first bit, but stomped on showing $pr^{3}-qs^{3}=0$

polar · « **Reply #1 on:** December 09, 2012, 04:08:55 pm »

$\begin{aligned} px^3 + qx^2 + rx +s &= 0 \\ x^3 + \frac{q}{p}x^2 + \frac{r}{p}x + \frac{s}{p} &= 0 \\ (x-\frac{a}{c})(x-a)(x-ca) &= x^3 - (ca + \frac{a}{c} - a)x^2 + (ca^2 + \frac{a^2}{c} - a^3)x - a^3 \\ x^3 + \frac{q}{p}x^2 } + \frac{r}{p}x + \frac{s}{p} &= x^3 - (ca + \frac{a}{c} - a)x^2 + (ca^2 + \frac{a^2}{c} - a^3)x - a^3 \\ \frac{s}{p} &= -a^3 \\ a &= \left(-\frac{s}{p}\right)^{\frac{1}{3}} \\ \\ \text{Since} \; x-a \; \text{is a factor of} \; P(x),} \; P(a) &=0 \\ P\left(\left(-\frac{s}{p}\right)^{\frac{1}{3}}\right) &=0 \\ \left(\left(-\frac{s}{p}}\right)^{\frac{1}{3}}\right)^3 + \frac{q}{p}\times\left(\left(-\frac{s}{p}\right)^{\frac{1}{3}}\right)^2 + \frac{r}{p}\times\left(-\frac{s}{p}\right)^\frac{1}{3} + \frac{s}{p} &=0 \\ -\frac{s}{p} + \frac{q}{p}\times\left(\frac{s}{p}\right)^{\frac{2}{3}} - \frac{r}{p}\times\left(\frac{s}{p}\right)^{\frac{1}{3}} + \frac{s}{p} &= 0 \\ \frac{q}{p}\times\left(\frac{s}{p}\right)^{\frac{2}{3}} &= \frac{r}{p}\times\left(\frac{s}{p}\right)^{\frac{1}{3}} \\ \frac{q}{r} &= \left(\frac{s}{p}\right)^{\frac{1}{3}} \times \left(\frac{p}{s}\right)^{\frac{2}{3}} \\ \frac{q}{r} &= \left(\frac{p}{s}\right)^{\frac{1}{3}} \\ \frac{q^3}{r^3} &= \frac{p}{s} \\ q^3s &= r^3p \\ pr^3 - q^3s &=0 \end{aligned}$

$pr^3 - qs^3$ should be $pr^3 - q^3s$ , if you try the polynomial $(x-1)(x-3)(x-9) = x^3 - 13x^2 + 39x -27$ , then, $pr^3 - q^3s = (1)(39)^3 - (-13)^3(-27) =0$ whereas the other one gave -196560

Nagisa · « **Reply #2 on:** December 09, 2012, 04:54:52 pm »

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.

polar · « **Reply #3 on:** December 09, 2012, 05:02:05 pm »

yeah lol, I thought if I did \left( and \right), the lines would be too long, I shall fix that now

Nagisa · « **Reply #4 on:** December 09, 2012, 05:03:41 pm »

Hancock · « **Reply #5 on:** December 09, 2012, 05:19:03 pm »

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.

Nagisa · « **Reply #6 on:** December 09, 2012, 05:19:56 pm »

Quote from: polar on December 09, 2012, 04:08:55 pm

$pr^3 - qs^3$ should be $pr^3 - q^3s$ , if you try the polynomial $(x-1)(x-3)(x-9) = x^3 - 13x^2 + 39x -27$ , then, $pr^3 - q^3s = (1)(39)^3 - (-13)^3(-27) =0$ 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

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Author Topic: Halp pls (Read 2800 times) Tweet Share

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