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Question Number 186571 by ajfour last updated on 06/Feb/23

If x=−(1/((1+m)))((2/3)±(√((4/9)±((ia(√m))/(12))))) where (1+m)^2 =am and that 9a(a+16)^2 =(16)^3 Find real x.

$${If}\:\:\:{x}=−\frac{\mathrm{1}}{\left(\mathrm{1}+{m}\right)}\left(\frac{\mathrm{2}}{\mathrm{3}}\pm\sqrt{\frac{\mathrm{4}}{\mathrm{9}}\pm\frac{{ia}\sqrt{{m}}}{\mathrm{12}}}\right) \\ $$$${where}\:\:\:\left(\mathrm{1}+{m}\right)^{\mathrm{2}} ={am} \\ $$$$\:\:\:\:{and}\:{that}\:\:\mathrm{9}{a}\left({a}+\mathrm{16}\right)^{\mathrm{2}} =\left(\mathrm{16}\right)^{\mathrm{3}} \\ $$$${Find}\:{real}\:{x}. \\ $$

Answered by ajfour last updated on 06/Feb/23

let a+16=(1/b) 9((1/b)−16)=(16)^3 b^2 9(1−16b)=(16)^3 b^3 (16b)^3 +9(16b)−9=0 16b=(((3(√(21))+9)/2))^(1/3) −(((3(√(21))−9)/2))^(1/3) a=((16)/((((3(√(21))+9)/2))^(1/3) −(((3(√(21))−9)/2))^(1/3) ))−16 a≈1.48810019 now m^2 +(2−a)m+1=0 m=(a/2)−1±(√((a^2 /4)−a)) m=(a/2)−1+i(√(a(1−(a/4))))=h+ik As a(1−(a/4))>0 m(∈C)=h+ik =(a/2)−1+i(√(a−(a^2 /4))) x=−(2/(3(1+h+ik))){1+(√(1±((3a(√(h+ik)))/(16))))} but m^(1/2) =(√(h+ik))=((1+h+ik)/a) let 16+3(1+h+ik)=(p+iq)^2 19+3h=p^2 −q^2 3k=2pq (19+3h)^2 =(p+q)^2 {(p+q)^2 −6k} (p+q)^3 −6k(p+q)^2 −(19+3h)^2 =0 say p+q=z z^3 −6(√(a(1−(a/4))))z^2 −(((3a)/2)+16)^2 =0 we get z>0 from above p,q =(z/2)±(√((z^2 /4)−(3/2)(√(a(1−(a/4)))))) hopefully p,q ∈R x=(2/(3(1+h+ik)))(1+((p+iq)/4)) x=((p+4+iq)/(6(1+h+ik)))×(((1+h−ik))/((1+h−ik))) x=(((p+4)(1+h)+kq)/(6{(1+h)^2 +k^2 })) +((i{q(1+h)−k(p+4)})/(6{(1+h)^2 +k^) })) If x has to ∈R ⇒ ((p+4)/q)=((1+h)/k)=((a/2)/( (√(a(1−(a/4)))))) p+4=((aq)/( (√(a(4−a))))) 4+(z/2)+(√((z^2 /4)−(3/2)(√(a(1−(a/4)))))) =(a/( (√(a(4−a))))){(z/2)−(√((z^2 /4)−(3/2)(√(a(1−(a/4))))))} ⇒ ((M−1)(z/2)−4)^2 =(M+1)^2 Q(z) .....

$${let}\:\:{a}+\mathrm{16}=\frac{\mathrm{1}}{{b}} \\ $$$$\mathrm{9}\left(\frac{\mathrm{1}}{{b}}−\mathrm{16}\right)=\left(\mathrm{16}\right)^{\mathrm{3}} {b}^{\mathrm{2}} \\ $$$$\mathrm{9}\left(\mathrm{1}−\mathrm{16}{b}\right)=\left(\mathrm{16}\right)^{\mathrm{3}} {b}^{\mathrm{3}} \\ $$$$\left(\mathrm{16}{b}\right)^{\mathrm{3}} +\mathrm{9}\left(\mathrm{16}{b}\right)−\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{16}{b}=\left(\frac{\mathrm{3}\sqrt{\mathrm{21}}+\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} −\left(\frac{\mathrm{3}\sqrt{\mathrm{21}}−\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${a}=\frac{\mathrm{16}}{\left(\frac{\mathrm{3}\sqrt{\mathrm{21}}+\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} −\left(\frac{\mathrm{3}\sqrt{\mathrm{21}}−\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} }−\mathrm{16} \\ $$$${a}\approx\mathrm{1}.\mathrm{48810019} \\ $$$${now}\:\:\:{m}^{\mathrm{2}} +\left(\mathrm{2}−{a}\right){m}+\mathrm{1}=\mathrm{0} \\ $$$${m}=\frac{{a}}{\mathrm{2}}−\mathrm{1}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{a}} \\ $$$${m}=\frac{{a}}{\mathrm{2}}−\mathrm{1}+{i}\sqrt{{a}\left(\mathrm{1}−\frac{{a}}{\mathrm{4}}\right)}={h}+{ik} \\ $$$${As}\:\:{a}\left(\mathrm{1}−\frac{{a}}{\mathrm{4}}\right)>\mathrm{0}\:\: \\ $$$${m}\left(\in\mathbb{C}\right)={h}+{ik}\:=\frac{{a}}{\mathrm{2}}−\mathrm{1}+{i}\sqrt{{a}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${x}=−\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{1}+{h}+{ik}\right)}\left\{\mathrm{1}+\sqrt{\mathrm{1}\pm\frac{\mathrm{3}{a}\sqrt{{h}+{ik}}}{\mathrm{16}}}\right\} \\ $$$${but}\:\:{m}^{\mathrm{1}/\mathrm{2}} =\sqrt{{h}+{ik}}=\frac{\mathrm{1}+{h}+{ik}}{{a}} \\ $$$${let}\:\:\mathrm{16}+\mathrm{3}\left(\mathrm{1}+{h}+{ik}\right)=\left({p}+{iq}\right)^{\mathrm{2}} \\ $$$$\mathrm{19}+\mathrm{3}{h}={p}^{\mathrm{2}} −{q}^{\mathrm{2}} \\ $$$$\mathrm{3}{k}=\mathrm{2}{pq} \\ $$$$\left(\mathrm{19}+\mathrm{3}{h}\right)^{\mathrm{2}} =\left({p}+{q}\right)^{\mathrm{2}} \left\{\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{6}{k}\right\} \\ $$$$\left({p}+{q}\right)^{\mathrm{3}} −\mathrm{6}{k}\left({p}+{q}\right)^{\mathrm{2}} −\left(\mathrm{19}+\mathrm{3}{h}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${say}\:\:{p}+{q}={z} \\ $$$${z}^{\mathrm{3}} −\mathrm{6}\sqrt{{a}\left(\mathrm{1}−\frac{{a}}{\mathrm{4}}\right)}{z}^{\mathrm{2}} −\left(\frac{\mathrm{3}{a}}{\mathrm{2}}+\mathrm{16}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${we}\:{get}\:\:{z}>\mathrm{0}\:\:\:{from}\:{above} \\ $$$${p},{q}\:=\frac{{z}}{\mathrm{2}}\pm\sqrt{\frac{{z}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{a}\left(\mathrm{1}−\frac{{a}}{\mathrm{4}}\right)}} \\ $$$${hopefully}\:\:{p},{q}\:\in\mathbb{R} \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{1}+{h}+{ik}\right)}\left(\mathrm{1}+\frac{{p}+{iq}}{\mathrm{4}}\right) \\ $$$${x}=\frac{{p}+\mathrm{4}+{iq}}{\mathrm{6}\left(\mathrm{1}+{h}+{ik}\right)}×\frac{\left(\mathrm{1}+{h}−{ik}\right)}{\left(\mathrm{1}+{h}−{ik}\right)} \\ $$$${x}=\frac{\left({p}+\mathrm{4}\right)\left(\mathrm{1}+{h}\right)+{kq}}{\mathrm{6}\left\{\left(\mathrm{1}+{h}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} \right\}}\:+\frac{{i}\left\{{q}\left(\mathrm{1}+{h}\right)−{k}\left({p}+\mathrm{4}\right)\right\}}{\mathrm{6}\left\{\left(\mathrm{1}+{h}\right)^{\mathrm{2}} +{k}^{\left.\right)} \right\}} \\ $$$${If}\:\:{x}\:{has}\:{to}\:\in\mathbb{R} \\ $$$$\Rightarrow\:\:\frac{{p}+\mathrm{4}}{{q}}=\frac{\mathrm{1}+{h}}{{k}}=\frac{{a}/\mathrm{2}}{\:\sqrt{{a}\left(\mathrm{1}−\frac{{a}}{\mathrm{4}}\right)}} \\ $$$${p}+\mathrm{4}=\frac{{aq}}{\:\sqrt{{a}\left(\mathrm{4}−{a}\right)}} \\ $$$$\mathrm{4}+\frac{{z}}{\mathrm{2}}+\sqrt{\frac{{z}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{a}\left(\mathrm{1}−\frac{{a}}{\mathrm{4}}\right)}} \\ $$$$=\frac{{a}}{\:\sqrt{{a}\left(\mathrm{4}−{a}\right)}}\left\{\frac{{z}}{\mathrm{2}}−\sqrt{\frac{{z}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}\sqrt{{a}\left(\mathrm{1}−\frac{{a}}{\mathrm{4}}\right)}}\right\} \\ $$$$\Rightarrow\:\:\left(\left({M}−\mathrm{1}\right)\frac{{z}}{\mathrm{2}}−\mathrm{4}\right)^{\mathrm{2}} =\left({M}+\mathrm{1}\right)^{\mathrm{2}} {Q}\left({z}\right) \\ $$$$..... \\ $$$$ \\ $$

Answered by Frix last updated on 06/Feb/23

(√(u+vi))∈R ⇒ v=0 ⇒ a(√m)=0 ⇔ a=0∨m=0 a=0 ⇒ 9a(a+16)^2 =16^3 ⇔ 0=16^3 is false m=0 ⇒ (1+m)^2 =am ⇔ 1=0 is false ⇒ no real x

$$\sqrt{{u}+{v}\mathrm{i}}\in\mathbb{R}\:\Rightarrow\:{v}=\mathrm{0} \\ $$$$\Rightarrow\:{a}\sqrt{{m}}=\mathrm{0}\:\Leftrightarrow\:{a}=\mathrm{0}\vee{m}=\mathrm{0} \\ $$$${a}=\mathrm{0}\:\Rightarrow\:\mathrm{9}{a}\left({a}+\mathrm{16}\right)^{\mathrm{2}} =\mathrm{16}^{\mathrm{3}} \:\Leftrightarrow\:\mathrm{0}=\mathrm{16}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{false} \\ $$$${m}=\mathrm{0}\:\Rightarrow\:\left(\mathrm{1}+{m}\right)^{\mathrm{2}} ={am}\:\Leftrightarrow\:\mathrm{1}=\mathrm{0}\:\mathrm{is}\:\mathrm{false} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{real}\:{x} \\ $$

Commented by ajfour last updated on 06/Feb/23

now i think MjS≠Frix

$${now}\:\:{i}\:{think}\:\:{MjS}\neq{Frix} \\ $$

Commented by Rasheed.Sindhi last updated on 07/Feb/23

Please keep looking in the mirror until you see Frix opposite to you without any plus, minus. :)

$${Please}\:{keep}\:{looking}\:{in}\:{the}\:{mirror} \\ $$$${until}\:{you}\:{see}\:{Frix}\:{opposite}\:{to}\:{you} \\ $$$$\left.{without}\:{any}\:{plus},\:{minus}.\::\right) \\ $$

Commented by MJS_new last updated on 07/Feb/23

I just looked in the mirror: I′m still very close to the one I have been yesterday and the day before yesterday. but maybe (1) MJS+sunday+monday=Frix is true. in this case find the value of (2) (Frix+2tuesdays)×(MJS−wednesday)

$$\mathrm{I}\:\mathrm{just}\:\mathrm{looked}\:\mathrm{in}\:\mathrm{the}\:\mathrm{mirror}:\:\mathrm{I}'\mathrm{m}\:\mathrm{still}\:\mathrm{very} \\ $$$$\mathrm{close}\:\mathrm{to}\:\mathrm{the}\:\mathrm{one}\:\mathrm{I}\:\mathrm{have}\:\mathrm{been}\:\mathrm{yesterday}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{day}\:\mathrm{before}\:\mathrm{yesterday}.\:\mathrm{but}\:\mathrm{maybe} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{MJS}+\mathrm{sunday}+\mathrm{monday}=\mathrm{Frix} \\ $$$$\mathrm{is}\:\mathrm{true}.\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{Frix}+\mathrm{2tuesdays}\right)×\left(\mathrm{MJS}−\mathrm{wednesday}\right) \\ $$

Commented by Frix last updated on 07/Feb/23

The answer is 42 as always.

$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42}\:\mathrm{as}\:\mathrm{always}. \\ $$

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