Polynomials (1 Viewer)

Casper_18 · Jun 30, 2009

The equation $x^{3} + 3px^{2} + 3qx + r = 0$ , where $p^{2} \neq q$ , has a double root. show that $4(p^{2} - q)(q^{2} - pr) = (pq - r)^{2}$
The equation $P(x)= x^{4} - px^{3} + qx^{2} - rx + s$ has roots $\alpha , \beta , \gamma ,\delta$ . show that:

(A) if $\alpha \beta = \gamma \delta$ , then $r^{2} = sp^{2}$

(B) if $\alpha + \beta = \gamma + \delta$ , then $p^{3} - 4pq + 8r$

gurmies · Jul 1, 2009

Question 2

(A)

$x^{4}-px^{3}+qx^{2}-rx+s=0 \\\\ \sum \alpha _{i}=\alpha+\beta+\gamma+\delta=p \\\\ \sum \alpha\beta\gamma_{i}=\alpha\beta\gamma+\alpha\beta\delta+\beta\gamma\delta+\alpha\gamma\delta=\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta)\\\\=\alpha\beta(\gamma+\delta)+\alpha\beta(\alpha+\beta)[Since\,\,\alpha\beta=\gamma\delta] \\\\ =\alpha\beta(\alpha+\beta+\gamma+\delta)=p(\alpha\beta)=r \\\\ \prod \alpha_{i}=\alpha\beta\gamma\delta=(\alpha\beta)^{2}=s \\\\ \therefore \alpha\beta = \sqrt{s} \\\\ \therefore p\sqrt{s}=r \Rightarrow r^{2}=sp^{2}$

(B)

$\sum \alpha _{i}=\alpha+\beta+\gamma+\delta=2(\alpha+\beta)=p \Rightarrow \alpha+\beta=\frac{p}{2} \\\\ \sum \alpha\beta_{i}=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\alpha(\gamma+\delta)+\beta(\gamma+\delta)+\alpha\beta+\gamma\delta \\\\ = (\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta=(\alpha+\beta)^{2}+\alpha\beta+\gamma\delta=q \Rightarrow \alpha\beta+\gamma\delta=q-\frac{p^{2}}{4} \\\\ \sum \alpha\beta\gamma_{i}=\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=\alpha\beta(\gamma+\delta)+\gamma\delta(\alpha+\beta) \\\\ = \alpha\beta(\alpha+\beta)+\gamma\delta(\alpha+\beta)=(\alpha+\beta)(\alpha\beta+\gamma\delta)=r \Rightarrow (\frac{p}{2})(q-\frac{p^{2}}{4})=r \\\\ \therefore p^{3}-4pq+8r=0$

Casper_18 · Jul 1, 2009

Thankyou, anyone have a clue for question 1??

Trebla · Jul 1, 2009

Casper_18 · Jul 2, 2009

Thanks a heap. I never would have thought to use the expression for x^2.

Timothy.Siu · Jul 2, 2009

yeah they put this in our first task for ext2, i had no idea how to do it, but i see its just heavy algebra now

Casper_18 · Jul 2, 2009

Timothy.Siu said:
yeah they put this in our first task for ext2, i had no idea how to do it, but i see its just heavy algebra now

Yeah, it is a hard question. I suppose I just have to do enough of them so I know how to do as many as possible automatically

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