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Question-219668

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Question Number 219668 by Rojarani last updated on 30/Apr/25
Answered by SdC355 last updated on 30/Apr/25
p=p(4−p)(4−p(4−p)) 1=(4−p)(p^2 −4p+4) 4p^2 −16p+16−p^3 +4p^2 −4p=1 p^3 −8p^2 +20p−15=0 p=3 , p=(5/2)±((√5)/2) ∴p=3 , q=3 r=3
$${p}={p}\left(\mathrm{4}−{p}\right)\left(\mathrm{4}−{p}\left(\mathrm{4}−{p}\right)\right) \\ $$$$\mathrm{1}=\left(\mathrm{4}−{p}\right)\left({p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{4}\right) \\ $$$$\mathrm{4}{p}^{\mathrm{2}} −\mathrm{16}{p}+\mathrm{16}−{p}^{\mathrm{3}} +\mathrm{4}{p}^{\mathrm{2}} −\mathrm{4}{p}=\mathrm{1} \\ $$$${p}^{\mathrm{3}} −\mathrm{8}{p}^{\mathrm{2}} +\mathrm{20}{p}−\mathrm{15}=\mathrm{0} \\ $$$${p}=\mathrm{3}\:,\:{p}=\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$$$\therefore{p}=\mathrm{3}\:,\:{q}=\mathrm{3}\:{r}=\mathrm{3} \\ $$
Commented by Ghisom last updated on 30/Apr/25
wrong. distinct real numbers
$$\mathrm{wrong}. \\ $$$$\mathrm{distinct}\:\mathrm{real}\:\mathrm{numbers} \\ $$
Commented by SdC355 last updated on 30/Apr/25
Opps.... i didn′t read ′′distinct number′′
$$\mathrm{Opps}….\:\mathrm{i}\:\mathrm{didn}'\mathrm{t}\:\mathrm{read}\:''\mathrm{distinct}\:\mathrm{number}'' \\ $$
Answered by Ghisom last updated on 30/Apr/25
r=q(4−q) q=p(4−p) ⇒ p(p−3)(p^6 −13p^5 +65p^4 −157p^3 +189p^2 −105p+21)=0 1. p=q=r=0 2. p=q=r=3 3. p^6 −13p^5 +65p^4 −157p^3 +189p^2 −105p+21=0 this can be factorized (p^3 −7p^2 +14p−7)(p^3 −6p^2 +9p−3)=0 because the variables in given equations are interchangeable we don′t have to solve for p. the solutions are p+q+r=6∨p+q+r=7
$${r}={q}\left(\mathrm{4}−{q}\right) \\ $$$${q}={p}\left(\mathrm{4}−{p}\right) \\ $$$$\Rightarrow \\ $$$${p}\left({p}−\mathrm{3}\right)\left({p}^{\mathrm{6}} −\mathrm{13}{p}^{\mathrm{5}} +\mathrm{65}{p}^{\mathrm{4}} −\mathrm{157}{p}^{\mathrm{3}} +\mathrm{189}{p}^{\mathrm{2}} −\mathrm{105}{p}+\mathrm{21}\right)=\mathrm{0} \\ $$$$\mathrm{1}. \\ $$$${p}={q}={r}=\mathrm{0} \\ $$$$\mathrm{2}. \\ $$$${p}={q}={r}=\mathrm{3} \\ $$$$\mathrm{3}. \\ $$$${p}^{\mathrm{6}} −\mathrm{13}{p}^{\mathrm{5}} +\mathrm{65}{p}^{\mathrm{4}} −\mathrm{157}{p}^{\mathrm{3}} +\mathrm{189}{p}^{\mathrm{2}} −\mathrm{105}{p}+\mathrm{21}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factorized} \\ $$$$\left({p}^{\mathrm{3}} −\mathrm{7}{p}^{\mathrm{2}} +\mathrm{14}{p}−\mathrm{7}\right)\left({p}^{\mathrm{3}} −\mathrm{6}{p}^{\mathrm{2}} +\mathrm{9}{p}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{because}\:\mathrm{the}\:\mathrm{variables}\:\mathrm{in}\:\mathrm{given}\:\mathrm{equations} \\ $$$$\mathrm{are}\:\mathrm{interchangeable}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{for}\:{p}.\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{are} \\ $$$${p}+{q}+{r}=\mathrm{6}\vee{p}+{q}+{r}=\mathrm{7} \\ $$
Commented by Ghisom last updated on 01/May/25
bonus content: x=y(k−y) y=z(k−z) z=x(k−x) trivial solutions x=y=z=0 x=y=z=k−1 the above method leads to x^6 −(3k+1)x^5 +(3k^2 +4k+1)x^4 −(k^3 +5k^2 +3k+1)x^3 +(2k^3 +3k^2 +3k+1)x^2 −(k^3 +2k^2 +2k+1)x+(k^2 +k+1)=0 we can also assume x+y+z=n_1 ∨x+y+z=n_2 which leads to Π_(j=1) ^2 (x^3 −n_j x^2 −(k^2 −(n_j −1)k−n_j −1)x+(((k−n_j +1)n_j )/2))=0 by expanding this one and matching with the above one we get n_1 =((3k+1−(√(k^2 −2k−7)))/2) n_2 =((3k+1+(√(k^2 −2k−7)))/2) ⇒ x^6 −(3k+1)x^5 +(3k^2 +4k+1)x^4 −(k^3 +5k^2 +3k+1)x^3 +(2k^3 +3k^2 +3k+1)x^2 −(k^3 +2k^2 +2k+1)x+(k^2 +k+1)= [let (√(k^2 −2k−7))=℧ =(x^3 −((3k+1+℧)/2)x^2 +((k^2 +2k−1+(k+1)℧)/2)x−((k^2 −k−2+k℧)/2))(x^3 −((3k+1−℧)/2)x^2 +((k^2 +2k−1−(k+1)℧)/2)x−((k^2 −k−2−k℧)/2)) ⇒ x+y+z ∈R ⇔ k≤1−2(√2)∨1+2(√2)≤k
$$\mathrm{bonus}\:\mathrm{content}: \\ $$$${x}={y}\left({k}−{y}\right) \\ $$$${y}={z}\left({k}−{z}\right) \\ $$$${z}={x}\left({k}−{x}\right) \\ $$$$\mathrm{trivial}\:\mathrm{solutions} \\ $$$${x}={y}={z}=\mathrm{0} \\ $$$${x}={y}={z}={k}−\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{above}\:\mathrm{method}\:\mathrm{leads}\:\mathrm{to}\: \\ $$$${x}^{\mathrm{6}} −\left(\mathrm{3}{k}+\mathrm{1}\right){x}^{\mathrm{5}} +\left(\mathrm{3}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}\right){x}^{\mathrm{4}} −\left({k}^{\mathrm{3}} +\mathrm{5}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right){x}^{\mathrm{3}} +\left(\mathrm{2}{k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right){x}^{\mathrm{2}} −\left({k}^{\mathrm{3}} +\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right){x}+\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{assume} \\ $$$${x}+{y}+{z}={n}_{\mathrm{1}} \vee{x}+{y}+{z}={n}_{\mathrm{2}} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\mathrm{2}} {\prod}}\left({x}^{\mathrm{3}} −{n}_{{j}} {x}^{\mathrm{2}} −\left({k}^{\mathrm{2}} −\left({n}_{{j}} −\mathrm{1}\right){k}−{n}_{{j}} −\mathrm{1}\right){x}+\frac{\left({k}−{n}_{{j}} +\mathrm{1}\right){n}_{{j}} }{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{by}\:\mathrm{expanding}\:\mathrm{this}\:\mathrm{one}\:\mathrm{and}\:\mathrm{matching}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{above}\:\mathrm{one}\:\mathrm{we}\:\mathrm{get} \\ $$$${n}_{\mathrm{1}} =\frac{\mathrm{3}{k}+\mathrm{1}−\sqrt{{k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{7}}}{\mathrm{2}} \\ $$$${n}_{\mathrm{2}} =\frac{\mathrm{3}{k}+\mathrm{1}+\sqrt{{k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{7}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{6}} −\left(\mathrm{3}{k}+\mathrm{1}\right){x}^{\mathrm{5}} +\left(\mathrm{3}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}\right){x}^{\mathrm{4}} −\left({k}^{\mathrm{3}} +\mathrm{5}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right){x}^{\mathrm{3}} +\left(\mathrm{2}{k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right){x}^{\mathrm{2}} −\left({k}^{\mathrm{3}} +\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right){x}+\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)= \\ $$$$\:\:\:\:\:\left[\mathrm{let}\:\sqrt{{k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{7}}=\mho\right. \\ $$$$=\left({x}^{\mathrm{3}} −\frac{\mathrm{3}{k}+\mathrm{1}+\mho}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{{k}^{\mathrm{2}} +\mathrm{2}{k}−\mathrm{1}+\left({k}+\mathrm{1}\right)\mho}{\mathrm{2}}{x}−\frac{{k}^{\mathrm{2}} −{k}−\mathrm{2}+{k}\mho}{\mathrm{2}}\right)\left({x}^{\mathrm{3}} −\frac{\mathrm{3}{k}+\mathrm{1}−\mho}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{{k}^{\mathrm{2}} +\mathrm{2}{k}−\mathrm{1}−\left({k}+\mathrm{1}\right)\mho}{\mathrm{2}}{x}−\frac{{k}^{\mathrm{2}} −{k}−\mathrm{2}−{k}\mho}{\mathrm{2}}\right) \\ $$$$\Rightarrow \\ $$$${x}+{y}+{z}\:\in\mathbb{R}\:\Leftrightarrow\:{k}\leqslant\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\vee\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\leqslant{k} \\ $$

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