p^3 +q^3 +p^2 +q^2 +c^2 =0 find p+q in terms of c. if c^2 <(4/9).


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 175462 by ajfour last updated on 30/Aug/22

$p^{3} + q^{3} + p^{2} + q^{2} + c^{2} = 0$ $f i n d p + q i n t e r m s o f c . i f c^{2} < \frac{4}{9} .$
Commented bymr W last updated on 01/Sep/22

$i t h i n k t h e r e i s n o u n i q u e v a l u e p + q,$ $b u t t h e r e a r e {(p + q)}_{m i n} a n d {(p + q)}_{m a x} .$
Commented byajfour last updated on 01/Sep/22

$v a l u e s m a y b e m u l t i p l e b u t a r e$ $c o n s t a n t s, i t h i n k .$
Commented bymr W last updated on 01/Sep/22

$f o r p, q \in R$ $- \frac{2}{\sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 + \frac{8}{27 c^{2}}} + 1)} - \sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 + \frac{8}{27 c^{2}}} - 1)}} ⩽ p + q < - \frac{2}{3}$
	Answered by mr W last updated on 04/Sep/22

	$l e t s = p + q$ $s^{3} - 3 s p q + s^{2} - 2 p q + c^{2} = 0$ $s^{3} - 3 s p (s - p) + s^{2} - 2 p (s - p) + c^{2} = 0$ $(3 s + 2) p^{2} - s (3 s + 2) p + s^{3} + s^{2} + c^{2} = 0$ $Δ = s^{2} {(3 s + 2)}^{2} - 4 (3 s + 2) (s^{3} + s^{2} + c^{2}) ⩾ 0$ $(3 s + 2) (s^{3} + 2 s^{2} + 4 c^{2}) ⩽ 0$ $\Rightarrow 3 s + 2 < 0 \Rightarrow s < - \frac{2}{3}$ $\Rightarrow s^{3} + 2 s^{2} + 4 c^{2} ⩾ 0 \Rightarrow s ⩾ - \frac{2}{\sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 + \frac{8}{27 c^{2}}} + 1)} - \sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 + \frac{8}{27 c^{2}}} - 1)}}$ $i . e . - \frac{2}{\sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 + \frac{8}{27 c^{2}}} + 1)} - \sqrt[3]{\frac{1}{c^{2}} (\sqrt{1 + \frac{8}{27 c^{2}}} - 1)}} ⩽ p + q < - \frac{2}{3}$
	Commented byTawa11 last updated on 15/Sep/22

	$Great sir .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum