3s^2 −2ps−3cp−1=0 and 3s−2p−sp^2 −3cp^2 =0 find s and p both real in terms of c ∈R.


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Question Number 79256 by ajfour last updated on 23/Jan/20

$3 s^{2} - 2 p s - 3 c p - 1 = 0 a n d$ $3 s - 2 p - {s p}^{2} - 3 {c p}^{2} = 0$ $f i n d s a n d p b o t h r e a l i n t e r m s$ $o f c \in R .$
	Answered by MJS last updated on 23/Jan/20

	$(1) c = \frac{s^{2}}{p} - \frac{2 s}{3} - \frac{1}{3 p}$ $(2) c = \frac{s}{p^{2}} - \frac{s}{3} - \frac{2}{3 p}$ $\frac{s^{2}}{p} - \frac{2 s}{3} - \frac{1}{3 p} = \frac{s}{p^{2}} - \frac{s}{3} - \frac{2}{3 p}$ $3 {p s}^{2} - p^{2} s - 3 s + p = 0$ $s = \frac{p}{3} \lor s = \frac{1}{p}$ $c = - \frac{p^{2} + 3}{9 p} \lor c = \frac{1 - p^{2}}{p^{3}}$ $\Rightarrow$ $p^{2} + 9 p c + 3 = 0 \lor p^{3} + \frac{1}{c} p^{2} - \frac{1}{c} = 0$ $and both can be solved$
	Commented by mr W last updated on 24/Jan/20

	$m o r e t h a n f a n t a s t i c!$
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