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 105448 by bemath last updated on 29/Jul/20

Given { ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z)),((((4z^2 )/(4z^2 +1)) = x)) :} . find x+y+z ?

$$\mathcal{G}{iven}\:\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases}\:\:\:.\:{find} \\ $$$${x}+{y}+{z}\:?\: \\ $$

Commented by john santu last updated on 29/Jul/20

(1/x)+(1/y)+(1/z)= ((4x^2 +1)/(4x^2 ))+((4y^2 +1)/(4y^2 ))+((4z^2 +1)/(4z^2 )) (1/x)+(1/y)+(1/z)= 3+(1/(4x^2 ))+(1/(4y^2 ))+(1/(4z^2 )) (1/x)+(1/y)+(1/z)=3+((1/(2x)))^2 +((1/(2y)))^2 +((1/(2z)))^2 (1/x)+(1/y)+(1/z)=3+(1/4){((1/x))^2 +((1/y))^2 +((1/z))^2 } { ((set (1/x)=a, (1/y)=b, (1/z)=c {: (),() })) :} ⇔ a+b+c = 3 +(1/4){(a+b+c)^2 −2(ab+ac+bc)}

$$\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\:\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }+\frac{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{y}^{\mathrm{2}} }+\frac{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{z}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\:\mathrm{3}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{y}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{z}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\mathrm{3}+\left(\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}{y}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}{z}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} \right\} \\ $$$$\begin{cases}{{set}\:\frac{\mathrm{1}}{{x}}={a},\:\frac{\mathrm{1}}{{y}}={b},\:\frac{\mathrm{1}}{{z}}={c}\:\left.\begin{matrix}{}\\{}\end{matrix}\right\}}\end{cases} \\ $$$$\Leftrightarrow\:{a}+{b}+{c}\:=\:\mathrm{3}\:+\frac{\mathrm{1}}{\mathrm{4}}\left\{\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{ac}+{bc}\right)\right\} \\ $$$$ \\ $$

Commented by behi83417@gmail.com last updated on 29/Jul/20

x=y=z=0 and:4x^2 −4x+1=0⇒x=y=z=(1/2) if:x≠y≠z≠0⇒x=(1/(2a)),y=(1/(2b)),z=(1/(2c))⇒ { ((((4.(1/(4a^2 )))/(4.(1/(4a^2 ))+1))=(1/(2b))⇒ { (((1/(1+a^2 ))=(1/(2b)))),(((1/(1+b^2 ))=(1/(2c))⇒)) :})),() :} ⇒ { ((a^2 +1=2b)),((b^2 +1=2c)),((c^2 +1=2a)) :}⇒(a−1)^2 +(b−1)^2 +(c−1)^2 =0 ⇒a=b=c=1⇒x=y=z=(1/2)

$$\mathrm{x}=\mathrm{y}=\mathrm{z}=\mathrm{0}\:\mathrm{and}:\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{if}:\mathrm{x}\neq\mathrm{y}\neq\mathrm{z}\neq\mathrm{0}\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2a}},\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2b}},\mathrm{z}=\frac{\mathrm{1}}{\mathrm{2c}}\Rightarrow \\ $$$$\begin{cases}{\frac{\mathrm{4}.\frac{\mathrm{1}}{\mathrm{4a}^{\mathrm{2}} }}{\mathrm{4}.\frac{\mathrm{1}}{\mathrm{4a}^{\mathrm{2}} }+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2b}}\Rightarrow\begin{cases}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2b}}}\\{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2c}}\Rightarrow}\end{cases}}\\{}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}=\mathrm{2b}}\\{\mathrm{b}^{\mathrm{2}} +\mathrm{1}=\mathrm{2c}}\\{\mathrm{c}^{\mathrm{2}} +\mathrm{1}=\mathrm{2a}}\end{cases}\Rightarrow\left(\mathrm{a}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{c}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by ajfour last updated on 29/Jul/20

let (1/x)=p , (1/y)=q , (1/z)=r ⇒ (4/(4−p^2 ))=(1/q) , (4/(4((q/p))^2 +q^2 ))= (1/r) , (4/(4+r^2 ))=(1/p) ⇒ 4−p^2 =4q ...(i) 4((q/p))^2 +q^2 =4r ⇒ q^2 ((1/p^2 )+(1/4))=r ..(ii) 4+r^2 =4p ...(iii) ⇒ (1−(p^2 /4))^2 ((1/p^2 )+(1/4))^2 =4(p−1) ⇒ (16−p^4 )^2 =64×16p^4 (p−1) ......degree 8

$${let}\:\:\:\frac{\mathrm{1}}{{x}}={p}\:\:\:,\:\:\frac{\mathrm{1}}{{y}}={q}\:\:,\:\:\frac{\mathrm{1}}{{z}}={r} \\ $$$$\Rightarrow\:\:\frac{\mathrm{4}}{\mathrm{4}−{p}^{\mathrm{2}} }=\frac{\mathrm{1}}{{q}}\:,\:\:\:\frac{\mathrm{4}}{\mathrm{4}\left(\frac{{q}}{{p}}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} }=\:\frac{\mathrm{1}}{{r}}\:, \\ $$$$\:\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{4}+{r}^{\mathrm{2}} }=\frac{\mathrm{1}}{{p}} \\ $$$$\Rightarrow\:\:\mathrm{4}−{p}^{\mathrm{2}} =\mathrm{4}{q}\:\:\:\:...\left({i}\right) \\ $$$$\:\mathrm{4}\left(\frac{{q}}{{p}}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{4}{r}\:\:\Rightarrow\:\:{q}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\right)={r}\:\:\:..\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{4}+{r}^{\mathrm{2}} =\mathrm{4}{p}\:\:\:\:\:\:\:...\left({iii}\right) \\ $$$$\Rightarrow\:\:\:\left(\mathrm{1}−\frac{{p}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{4}\left({p}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{16}−{p}^{\mathrm{4}} \right)^{\mathrm{2}} =\mathrm{64}×\mathrm{16}{p}^{\mathrm{4}} \left({p}−\mathrm{1}\right) \\ $$$$......{degree}\:\mathrm{8} \\ $$

Answered by bramlex last updated on 29/Jul/20

(1/y) = ((4x^2 +1)/(4x^2 )) →(1/y) = 1+(1/(4x^2 )) →(1/z) = 1+(1/4)(1+(1/(4x^2 )))^2 →1+(1/(4z^2 )) = (1/x) →1+(1/4){1+(1/4)(1+(1/(4x^2 )))^2 }= (1/x) set (1/x) = m 1+(1/4){1+(1/4)(1+(1/4)m^2 )^2 } = m (5/4)+(1/4)(1+(1/2)m^2 +(1/(16))m^4 )=m 20+4+2m^2 +(1/4)m^4 =16m m^4 +32m^2 −256m+16×24=0

$$\frac{\mathrm{1}}{{y}}\:=\:\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\:\rightarrow\frac{\mathrm{1}}{{y}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$\rightarrow\frac{\mathrm{1}}{{z}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\rightarrow\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{z}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{x}} \\ $$$$\rightarrow\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \right\}=\:\frac{\mathrm{1}}{{x}} \\ $$$${set}\:\frac{\mathrm{1}}{{x}}\:=\:{m}\: \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}{m}^{\mathrm{2}} \right)^{\mathrm{2}} \right\}\:=\:{m}\: \\ $$$$\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{m}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{16}}{m}^{\mathrm{4}} \right)={m} \\ $$$$\mathrm{20}+\mathrm{4}+\mathrm{2}{m}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{m}^{\mathrm{4}} =\mathrm{16}{m} \\ $$$${m}^{\mathrm{4}} +\mathrm{32}{m}^{\mathrm{2}} −\mathrm{256}{m}+\mathrm{16}×\mathrm{24}=\mathrm{0} \\ $$

Answered by 1549442205PVT last updated on 07/Aug/20

WLOG suppose that x≥y≥z≥0 (1).Then ((4z^2 )/(4z^2 +1))≥((4x^2 )/(4x^2 +1))≥((4y^2 )/(4y^2 +1))≥0 ⇒(z^2 /(4z^2 +1))≥(x^2 /(4x^2 +1))⇒z^2 (4x^2 +1)≥x^2 (4z^2 +1) ⇒z^2 ≥x^2 ⇒z≥x≥0(2).From (1) and(2) we get x=y=z.Hence ((4x^2 )/(4x^2 +1))=x⇔4x^3 −4x^2 +x=0 ⇔x(4x^2 −4x+1)=0.⇔x(2x−1)^2 =0 ⇔x∈{0;(1/2)} Thus,the roots of the given system are: {(0;0);((1/2);(1/2))}

$$\mathrm{WLOG}\:\mathrm{suppose}\:\mathrm{that}\:\mathrm{x}\geqslant\mathrm{y}\geqslant\mathrm{z}\geqslant\mathrm{0}\:\left(\mathrm{1}\right).\mathrm{Then} \\ $$$$\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{4z}^{\mathrm{2}} +\mathrm{1}}\geqslant\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{4x}^{\mathrm{2}} +\mathrm{1}}\geqslant\frac{\mathrm{4y}^{\mathrm{2}} }{\mathrm{4y}^{\mathrm{2}} +\mathrm{1}}\geqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{4z}^{\mathrm{2}} +\mathrm{1}}\geqslant\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4x}^{\mathrm{2}} +\mathrm{1}}\Rightarrow\mathrm{z}^{\mathrm{2}} \left(\mathrm{4x}^{\mathrm{2}} +\mathrm{1}\right)\geqslant\mathrm{x}^{\mathrm{2}} \left(\mathrm{4z}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{2}} \geqslant\mathrm{x}^{\mathrm{2}} \Rightarrow\mathrm{z}\geqslant\mathrm{x}\geqslant\mathrm{0}\left(\mathrm{2}\right).\mathrm{From}\:\left(\mathrm{1}\right)\:\mathrm{and}\left(\mathrm{2}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}=\mathrm{y}=\mathrm{z}.\mathrm{Hence} \\ $$$$\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{4x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{x}\Leftrightarrow\mathrm{4x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{x}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{x}\left(\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}\right)=\mathrm{0}.\Leftrightarrow\mathrm{x}\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\boldsymbol{\mathrm{x}}\in\left\{\mathrm{0};\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{system}\:\mathrm{are}: \\ $$$$\left\{\left(\mathrm{0};\mathrm{0}\right);\left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}}\right)\right\} \\ $$

Commented by behi83417@gmail.com last updated on 31/Jul/20

sir 154! there is a typo in line#5 from end.

$$\mathrm{sir}\:\mathrm{154}!\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{line}#\mathrm{5}\:\mathrm{from}\:\mathrm{end}. \\ $$

Commented by 1549442205PVT last updated on 01/Aug/20

Thank Sir.I understanded and shall correct it

$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{I}\:\mathrm{understanded}\:\mathrm{and}\:\mathrm{shall}\: \\ $$$$\mathrm{correct}\:\mathrm{it} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com