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Question Number 119997 by behi83417@gmail.com last updated on 28/Oct/20

solve for x,a∈R. (√(x^2 +ax+a^2 ))+(√(x^2 −ax+a^2 ))=1

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\mathrm{a}\in\boldsymbol{\mathrm{R}}. \\ $$$$\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{ax}}+\boldsymbol{{a}}^{\mathrm{2}} }+\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{ax}}+\boldsymbol{{a}}^{\mathrm{2}} }=\mathrm{1} \\ $$

Answered by Lordose last updated on 28/Oct/20

x^2 +ax+a^2 = 1 − 2(√(x^2 −ax+a^2 )) + x^2 −ax^2 +a^2 2(√(x^2 −ax+a^2 )) = 1−x^2 −ax−a^2 +x^2 −ax+a^2 4(x^2 −ax+a^2 ) = (1−2ax)^2 4x^2 −4ax+4a^2 = 1 − 4ax + 4a^2 x^2 (4−4a^2 )x^2 + 4a^2 = 1 x^2 = ((1−4a^2 )/((4−4a^2 ))) x = ±((√(1−4a^2 ))/(2(√(1−a^2 ))))

$$\mathrm{x}^{\mathrm{2}} +\mathrm{ax}+\mathrm{a}^{\mathrm{2}} \:=\:\mathrm{1}\:−\:\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{a}^{\mathrm{2}} }\:+\:\mathrm{x}^{\mathrm{2}} −\mathrm{ax}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{a}^{\mathrm{2}} }\:=\:\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{ax}−\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{a}^{\mathrm{2}} \\ $$$$\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{a}^{\mathrm{2}} \right)\:=\:\left(\mathrm{1}−\mathrm{2ax}\right)^{\mathrm{2}} \\ $$$$\mathrm{4x}^{\mathrm{2}} −\mathrm{4ax}+\mathrm{4a}^{\mathrm{2}} \:=\:\mathrm{1}\:−\:\mathrm{4ax}\:+\:\mathrm{4a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{4}−\mathrm{4a}^{\mathrm{2}} \right)\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4a}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{1}−\mathrm{4a}^{\mathrm{2}} }{\left(\mathrm{4}−\mathrm{4a}^{\mathrm{2}} \right)} \\ $$$$\mathrm{x}\:=\:\pm\frac{\sqrt{\mathrm{1}−\mathrm{4a}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }} \\ $$

Commented by behi83417@gmail.com last updated on 28/Oct/20

thank you very much sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}. \\ $$

Answered by TANMAY PANACEA last updated on 28/Oct/20

x^2 +ax+a^2 =1+x^2 −ax+a^2 −2(√(x^2 −ax+a^2 )) (2ax−1)^2 =4(x^2 −ax+a^2 ) 4a^2 x^2 −4ax+1−4x^2 +4ax−4a^2 =0 4x^2 (a^2 −1)=4a^2 −1 x=±(√((4a^2 −1)/(4(a^2 −1))))

$${x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} =\mathrm{1}+{x}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} −\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}{ax}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left({x}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} \right) \\ $$$$\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{4}{ax}+\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{ax}−\mathrm{4}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \left({a}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{4}{a}^{\mathrm{2}} −\mathrm{1} \\ $$$${x}=\pm\sqrt{\frac{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}}\: \\ $$

Commented by behi83417@gmail.com last updated on 28/Oct/20

thanks in advance sir!

$$\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}\:\mathrm{sir}! \\ $$

Commented by TANMAY PANACEA last updated on 28/Oct/20

most welcome sir

$${most}\:{welcome}\:{sir} \\ $$

Answered by behi83417@gmail.com last updated on 28/Oct/20

x^2 +ax+a^2 =u^2 ,x^2 −ax+a^2 =v^2 ⇒ { ((u+v=1)),((u^2 −v^2 =2ax)) :}⇒u−v=2ax ⇒(u+v)^2 −4uv=(2ax)^2 ⇒uv=((1−4a^2 x^2 )/4),u+v=1 ⇒z^2 −z+((1−4a^2 x^2 )/4)=0 ⇒z=((1±(√(1−(1−4a^2 x^2 ))))/2)=((1±2ax)/2) ⇒(√(x^2 +ax+a^2 ))=((1±2ax)/2)⇒ 4(x^2 +ax+a^2 )=1±4ax+4a^2 x^2 ⇒ { ((4(1−a^2 )x^2 =1−4a^2 ⇒x=±(1/2).(√((1−4a^2 )/(1−a^2 ))))),((4x^2 +4ax+4a^2 =1−4ax+4a^2 x^2 ⇒)) :} ⇒4(1−a^2 )x^2 +8ax+4a^2 −1=0⇒ ⇒x=((−4a±(√(16a^2 −4(1−a^2 )(4a^2 −1))))/(4(1−a^2 )))= =((−2a±(√(4a^2 −(4a^2 −1−4a^4 +a^2 ))))/(2(1−a^2 )))= =((2a±(√(4a^4 −a^2 +1)))/(2(a^2 −1))) .■

$$\mathrm{x}^{\mathrm{2}} +\mathrm{ax}+\mathrm{a}^{\mathrm{2}} =\mathrm{u}^{\mathrm{2}} ,\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{a}^{\mathrm{2}} =\mathrm{v}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{u}+\mathrm{v}=\mathrm{1}}\\{\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} =\mathrm{2ax}}\end{cases}\Rightarrow\mathrm{u}−\mathrm{v}=\mathrm{2ax} \\ $$$$\Rightarrow\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{2}} −\mathrm{4uv}=\left(\mathrm{2ax}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{uv}=\frac{\mathrm{1}−\mathrm{4a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }{\mathrm{4}},\mathrm{u}+\mathrm{v}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\frac{\mathrm{1}−\mathrm{4a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{z}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{4a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}}{\mathrm{2}}=\frac{\mathrm{1}\pm\mathrm{2ax}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{ax}+\mathrm{a}^{\mathrm{2}} }=\frac{\mathrm{1}\pm\mathrm{2ax}}{\mathrm{2}}\Rightarrow \\ $$$$\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{ax}+\mathrm{a}^{\mathrm{2}} \right)=\mathrm{1}\pm\mathrm{4ax}+\mathrm{4a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{4}\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\mathrm{x}^{\mathrm{2}} =\mathrm{1}−\mathrm{4a}^{\mathrm{2}} \Rightarrow\mathrm{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}.\sqrt{\frac{\mathrm{1}−\mathrm{4a}^{\mathrm{2}} }{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}}\\{\mathrm{4x}^{\mathrm{2}} +\mathrm{4ax}+\mathrm{4a}^{\mathrm{2}} =\mathrm{1}−\mathrm{4ax}+\mathrm{4a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \Rightarrow}\end{cases} \\ $$$$\Rightarrow\mathrm{4}\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\mathrm{x}^{\mathrm{2}} +\mathrm{8ax}+\mathrm{4a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$\Rightarrow\mathrm{x}=\frac{−\mathrm{4a}\pm\sqrt{\mathrm{16a}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{4a}^{\mathrm{2}} −\mathrm{1}\right)}}{\mathrm{4}\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)}= \\ $$$$=\frac{−\mathrm{2a}\pm\sqrt{\mathrm{4a}^{\mathrm{2}} −\left(\mathrm{4a}^{\mathrm{2}} −\mathrm{1}−\mathrm{4a}^{\mathrm{4}} +\mathrm{a}^{\mathrm{2}} \right)}}{\mathrm{2}\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)}= \\ $$$$=\frac{\mathrm{2a}\pm\sqrt{\mathrm{4a}^{\mathrm{4}} −\mathrm{a}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{1}\right)}\:\:\:\:.\blacksquare \\ $$

Answered by MJS_new last updated on 29/Oct/20

for a=0 we get x=±(1/2) for a=±(1/2) we get x=0 for a<−(1/2)∨a>(1/2) we get no solution for x∈C ⇒ −(1/2)≤a≤(1/2)∧−(1/2)≤x≤(1/2) ⇒ let a=(1/2)sin α ∧α∈R ⇒ x=±((cos α)/( (√(3+cos^2 α)))) parametric solution is { ((a=(1/2)sin α)),((x=±((cos α)/( (√(3+cos^2 α)))))) :}∧α∈R or with t=tan (α/2) { ((a=(t/(t^2 +1)))),((x=±((t^2 −1)/(2(√(t^4 +t^2 +1)))))) :}∧t∈R or solution is x=±((√(1−4a^2 ))/(2(√(1−a^2 ))))∧−(1/2)≤a≤(1/2)

$$\mathrm{for}\:{a}=\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:{x}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{for}\:{a}=\pm\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{we}\:\mathrm{get}\:{x}=\mathrm{0} \\ $$$$\mathrm{for}\:{a}<−\frac{\mathrm{1}}{\mathrm{2}}\vee{a}>\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{we}\:\mathrm{get}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$$\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{a}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\wedge−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{let}\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\alpha\:\wedge\alpha\in\mathbb{R} \\ $$$$\Rightarrow\:{x}=\pm\frac{\mathrm{cos}\:\alpha}{\:\sqrt{\mathrm{3}+\mathrm{cos}^{\mathrm{2}} \:\alpha}} \\ $$$$\mathrm{parametric}\:\mathrm{solution}\:\mathrm{is}\:\begin{cases}{{a}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\alpha}\\{{x}=\pm\frac{\mathrm{cos}\:\alpha}{\:\sqrt{\mathrm{3}+\mathrm{cos}^{\mathrm{2}} \:\alpha}}}\end{cases}\wedge\alpha\in\mathbb{R} \\ $$$$\mathrm{or}\:\mathrm{with}\:{t}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\begin{cases}{{a}=\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}}\\{{x}=\pm\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\sqrt{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}}}}\end{cases}\wedge{t}\in\mathbb{R} \\ $$$$\mathrm{or}\:\mathrm{solution}\:\mathrm{is}\:{x}=\pm\frac{\sqrt{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\wedge−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{a}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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

thank you very much dear proph:MJS please let me know what you think about my answer.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{proph}:\mathrm{MJS} \\ $$$$\mathrm{please}\:\mathrm{let}\:\mathrm{me}\:\mathrm{know}\:\mathrm{what}\:\mathrm{you}\:\mathrm{think} \\ $$$$\mathrm{about}\:\mathrm{my}\:\mathrm{answer}. \\ $$

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