Question-202500

Question Number 202500 by Calculusboy last updated on 28/Dec/23

Answered by Frix last updated on 28/Dec/23

Obviously x=1 ((1+3−3))^(1/(2015)) +((−1−3+5))^(1/(2015)) =1+1=2

$$\mathrm{Obviously}\:{x}=\mathrm{1} \\ $$$$\sqrt[{\mathrm{2015}}]{\mathrm{1}+\mathrm{3}−\mathrm{3}}+\sqrt[{\mathrm{2015}}]{−\mathrm{1}−\mathrm{3}+\mathrm{5}}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 28/Dec/23

((x^3 +3x−3))^(1/(2015)) +((−x^3 −3x+3+2))^(1/(2015)) =2 ((x^3 +3x−3))^(1/(2015)) +((−(x^3 +3x−3)+2))^(1/(2015)) =2 ▶((x^3 +3x−3))^(1/(2015)) =a⇒a^(2015) =x^3 +3x−3 ▶((−(x^3 +3x−3)+2))^(1/(2015)) =((2−a))^(1/(2015)) a+((2−a))^(1/(2015)) =2 ((2−a))^(1/(2015)) =2−a 2−a=(2−a)^(2015) (2−a)^(2015−1) =1 (2−a)^(2014) =1^(2014) 2−a=1 a=1 ((x^3 +3x−3))^(1/(2015)) =1 x^3 +3x−3=1 x^3 +3x−4=0 (x−1)(x^2 +x+4)=0 x=1 ∣ x=((−1±(√(1−16)))/2) x=1 ∣ x=((−1±i(√(15)))/2)

$$\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{2015}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{3}+\mathrm{2}}\:=\mathrm{2} \\ $$$$\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{2015}}]{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}\right)+\mathrm{2}}\:=\mathrm{2} \\ $$$$\blacktriangleright\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:={a}\Rightarrow{a}^{\mathrm{2015}} ={x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3} \\ $$$$\blacktriangleright\sqrt[{\mathrm{2015}}]{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}\right)+\mathrm{2}}\:=\sqrt[{\mathrm{2015}}]{\mathrm{2}−{a}}\: \\ $$$${a}+\sqrt[{\mathrm{2015}}]{\mathrm{2}−{a}}\:=\mathrm{2} \\ $$$$\sqrt[{\mathrm{2015}}]{\mathrm{2}−{a}}\:=\mathrm{2}−{a} \\ $$$$\:\:\:\:\:\mathrm{2}−{a}=\left(\mathrm{2}−{a}\right)^{\mathrm{2015}} \\ $$$$\:\left(\mathrm{2}−{a}\right)^{\mathrm{2015}−\mathrm{1}} =\mathrm{1} \\ $$$$\:\:\:\left(\mathrm{2}−{a}\right)^{\mathrm{2014}} =\mathrm{1}^{\mathrm{2014}} \\ $$$$\:\:\:\:\:\mathrm{2}−{a}=\mathrm{1} \\ $$$$\:\:\:\:\:\:{a}=\mathrm{1} \\ $$$$\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:=\mathrm{1} \\ $$$$\:\:\:\:\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}=\mathrm{1} \\ $$$$\:\:\:\:\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\:\:\:{x}=\mathrm{1}\:\mid\:{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{16}}}{\mathrm{2}} \\ $$$$\:\:\:{x}=\mathrm{1}\:\mid\:{x}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$

Commented by Rasheed.Sindhi last updated on 28/Dec/23

$${Erraneous}\:{answer}! \\ $$

Commented by Frix last updated on 28/Dec/23

I think we need (√1)+(√1)=2 ⇒ x^3 +3x−4=1 and your solution is right. I don′t think solutions for ((re^(iθ) ))^(1/n) +((2−re^(iθ) ))^(1/n) ∈R exist when re^(iθ) ∉R

$$\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{need}\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{1}}=\mathrm{2}\:\Rightarrow \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{right}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{solutions}\:\mathrm{for} \\ $$$$\sqrt[{{n}}]{{r}\mathrm{e}^{\mathrm{i}\theta} }+\sqrt[{{n}}]{\mathrm{2}−{r}\mathrm{e}^{\mathrm{i}\theta} }\in\mathbb{R} \\ $$$$\mathrm{exist}\:\mathrm{when}\:{r}\mathrm{e}^{\mathrm{i}\theta} \notin\mathbb{R} \\ $$

Commented by esmaeil last updated on 28/Dec/23

if(((x^3 +3x−3))^(1/(2015)) =a)→ ((2−(x^3 +3x−3)))^(1/(2015)) =((2−a^(2015) ))^(1/(2015)) ≠((2−a))^(1/(2015))

$${if}\left(\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}={a}\right)\rightarrow \\ $$$$\sqrt[{\mathrm{2015}}]{\mathrm{2}−\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}\right)}=\sqrt[{\mathrm{2015}}]{\mathrm{2}−{a}^{\mathrm{2015}} } \\ $$$$\neq\sqrt[{\mathrm{2015}}]{\mathrm{2}−{a}} \\ $$

Commented by Frix last updated on 28/Dec/23

But if x^3 +3x−3=a ⇒ (a)^(1/(2015)) +((2−a))^(1/(2015))

$$\mathrm{But}\:\mathrm{if}\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}={a}\:\Rightarrow\:\sqrt[{\mathrm{2015}}]{{a}}+\sqrt[{\mathrm{2015}}]{\mathrm{2}−{a}} \\ $$

Commented by esmaeil last updated on 28/Dec/23

$${yes} \\ $$$${but}\:{it}\:{dosn}'{t}\:{help}\:\:{us}\:{to}\:{solve}\:{that}. \\ $$

Commented by Frix last updated on 28/Dec/23

It helps to see there′s no other solutions besides a=1.

$$\mathrm{It}\:\mathrm{helps}\:\mathrm{to}\:\mathrm{see}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{other}\:\mathrm{solutions} \\ $$$$\mathrm{besides}\:{a}=\mathrm{1}. \\ $$

Commented by Calculusboy last updated on 28/Dec/23

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/23

$$\mathbb{T}\boldsymbol{\mathrm{hanks}}\:\boldsymbol{\mathrm{sir}}\:\mathrm{for}\:\boldsymbol{\mathrm{guidance}}! \\ $$

Answered by Rasheed.Sindhi last updated on 28/Dec/23

((x^3 +3x−3))^(1/(2015)) +((−(x^3 +3x−3)+2))^(1/(2015)) =2 (a)^(1/(2015)) +((2−a))^(1/(2015)) =2 A+B=2 A^(2015) +B^(2015) =a+2−a=2

$$\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{2015}}]{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}\right)+\mathrm{2}}\:=\mathrm{2} \\ $$$$\sqrt[{\mathrm{2015}}]{{a}}\:+\sqrt[{\mathrm{2015}}]{\mathrm{2}−{a}}\:=\mathrm{2} \\ $$$$\:\:\:{A}+{B}=\mathrm{2} \\ $$$${A}^{\mathrm{2015}} +{B}^{\mathrm{2015}} ={a}+\mathrm{2}−{a}=\mathrm{2} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 29/Dec/23

According to the strategy of sir Frix ((x^3 +3x−3))^(1/(2015)) +((−x^3 −3x+5))^(1/(2015)) ((x^3 +3x−4+1))^(1/(2015)) +((−(x^3 +3x−4)+1))^(1/(2015)) x^3 +3x−4=y ((1+y))^(1/(2015)) +((1−y))^(1/(2015)) =2 One possibility: ((1+y))^(1/(2015)) +((1−y))^(1/(2015)) =1+1 ((1+y))^(1/(2015)) =1 ∧ ((1−y))^(1/(2015)) =1 1+y=1^(2015) ∧ 1−y=1^(2015) 1+y=1 ∧ 1−y=1⇒y=0 x^3 +3x−4=0 (x−1)(x^2 +x+4)=0 x=1 , ((−1±i(√(15)) )/2)

$$\boldsymbol{\mathrm{According}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{strategy}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{Frix}} \\ $$$$\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{2015}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}} \\ $$$$\: \\ $$$$\sqrt[{\mathrm{2015}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}+\mathrm{1}}\:+\sqrt[{\mathrm{2015}}]{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}\right)+\mathrm{1}} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}={y} \\ $$$$\sqrt[{\mathrm{2015}}]{\mathrm{1}+{y}}\:+\sqrt[{\mathrm{2015}}]{\mathrm{1}−{y}}\:=\mathrm{2} \\ $$$${One}\:{possibility}: \\ $$$$\sqrt[{\mathrm{2015}}]{\mathrm{1}+{y}}\:+\sqrt[{\mathrm{2015}}]{\mathrm{1}−{y}}\:=\mathrm{1}+\mathrm{1} \\ $$$$\:\:\:\sqrt[{\mathrm{2015}}]{\mathrm{1}+{y}}\:=\mathrm{1}\:\wedge\:\sqrt[{\mathrm{2015}}]{\mathrm{1}−{y}}\:=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1}+{y}=\mathrm{1}^{\mathrm{2015}} \:\:\:\wedge\:\mathrm{1}−{y}=\mathrm{1}^{\mathrm{2015}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1}+{y}=\mathrm{1}\:\:\:\wedge\:\mathrm{1}−{y}=\mathrm{1}\Rightarrow{y}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{1}\:,\:\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{15}}\:}{\mathrm{2}} \\ $$

Commented by mr W last updated on 29/Dec/23

if you consider x∈C, i think there are much more complex roots than ((−1±i(√(15)))/2), because y=0 is the only one real root from ((1+y))^(1/(2015)) +((1−y))^(1/(2015)) =2, but not the only one complex root from it.

$${if}\:{you}\:{consider}\:{x}\in{C},\:{i}\:{think}\:{there}\: \\ $$$${are}\:{much}\:{more}\:{complex}\:{roots}\:{than} \\ $$$$\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{15}}}{\mathrm{2}},\:{because}\:{y}=\mathrm{0}\:{is}\:{the}\:{only}\: \\ $$$${one}\:{real}\:{root}\:{from}\:\sqrt[{\mathrm{2015}}]{\mathrm{1}+{y}}\:+\sqrt[{\mathrm{2015}}]{\mathrm{1}−{y}}\:=\mathrm{2}, \\ $$$${but}\:{not}\:{the}\:{only}\:{one}\:{complex}\:{root}\: \\ $$$${from}\:{it}. \\ $$

Commented by Frix last updated on 29/Dec/23

I don′t think ((1+y))^(1/n) +((1−y))^(1/n) =2 has complex roots.

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\sqrt[{{n}}]{\mathrm{1}+{y}}+\sqrt[{{n}}]{\mathrm{1}−{y}}=\mathrm{2}\:\mathrm{has}\:\mathrm{complex} \\ $$$$\mathrm{roots}. \\ $$

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