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Question Number 175462 by ajfour last updated on 30/Aug/22

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

$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{0} \\ $$ $${find}\:{p}+{q}\:{in}\:{terms}\:{of}\:{c}.\:\:{if}\:\:{c}^{\mathrm{2}} <\frac{\mathrm{4}}{\mathrm{9}}. \\ $$

Commented bymr W last updated on 01/Sep/22

i think there is no unique value p+q, but there are (p+q)_(min) and (p+q)_(max) .

$${i}\:{think}\:{there}\:{is}\:{no}\:{unique}\:{value}\:{p}+{q}, \\ $$ $${but}\:{there}\:{are}\:\left({p}+{q}\right)_{{min}} \:{and}\:\left({p}+{q}\right)_{{max}} . \\ $$

Commented byajfour last updated on 01/Sep/22

values may be multiple but are constants, i think.

$${values}\:{may}\:{be}\:{multiple}\:{but}\:{are} \\ $$ $${constants},\:{i}\:{think}. \\ $$

Commented bymr W last updated on 01/Sep/22

for p,q∈R −(2/( (((1/c^2 )((√(1+(8/(27c^2 ))))+1)))^(1/3) −(((1/c^2 )((√(1+(8/(27c^2 ))))−1)))^(1/3) ))≤p+q<−(2/3)

$${for}\:{p},{q}\in{R} \\ $$ $$−\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)}}\leqslant{p}+{q}<−\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Answered by mr W last updated on 04/Sep/22

let s=p+q s^3 −3spq+s^2 −2pq+c^2 =0 s^3 −3sp(s−p)+s^2 −2p(s−p)+c^2 =0 (3s+2)p^2 −s(3s+2)p+s^3 +s^2 +c^2 =0 Δ=s^2 (3s+2)^2 −4(3s+2)(s^3 +s^2 +c^2 )≥0 (3s+2)(s^3 +2s^2 +4c^2 )≤0 ⇒3s+2<0 ⇒s<−(2/3) ⇒s^3 +2s^2 +4c^2 ≥0 ⇒s≥−(2/( (((1/c^2 )((√(1+(8/(27c^2 ))))+1)))^(1/3) −(((1/c^2 )((√(1+(8/(27c^2 ))))−1)))^(1/3) )) i.e. −(2/( (((1/c^2 )((√(1+(8/(27c^2 ))))+1)))^(1/3) −(((1/c^2 )((√(1+(8/(27c^2 ))))−1)))^(1/3) ))≤p+q<−(2/3)

$${let}\:{s}={p}+{q} \\ $$ $${s}^{\mathrm{3}} −\mathrm{3}{spq}+{s}^{\mathrm{2}} −\mathrm{2}{pq}+{c}^{\mathrm{2}} =\mathrm{0} \\ $$ $${s}^{\mathrm{3}} −\mathrm{3}{sp}\left({s}−{p}\right)+{s}^{\mathrm{2}} −\mathrm{2}{p}\left({s}−{p}\right)+{c}^{\mathrm{2}} =\mathrm{0} \\ $$ $$\left(\mathrm{3}{s}+\mathrm{2}\right){p}^{\mathrm{2}} −{s}\left(\mathrm{3}{s}+\mathrm{2}\right){p}+{s}^{\mathrm{3}} +{s}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{0} \\ $$ $$\Delta={s}^{\mathrm{2}} \left(\mathrm{3}{s}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{3}{s}+\mathrm{2}\right)\left({s}^{\mathrm{3}} +{s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$ $$\left(\mathrm{3}{s}+\mathrm{2}\right)\left({s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \right)\leqslant\mathrm{0} \\ $$ $$\Rightarrow\mathrm{3}{s}+\mathrm{2}<\mathrm{0}\:\Rightarrow{s}<−\frac{\mathrm{2}}{\mathrm{3}} \\ $$ $$\Rightarrow{s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow{s}\geqslant−\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)}} \\ $$ $${i}.{e}.\:−\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)}}\leqslant{p}+{q}<−\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Commented byTawa11 last updated on 15/Sep/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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