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Question Number 177176 by mr W last updated on 01/Oct/22

solve (√(25−x^3 ))+(√(144−x^3 ))=13

$${solve} \\ $$$$\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }+\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }=\mathrm{13} \\ $$

Answered by a.lgnaoui last updated on 02/Oct/22

posons x^3 =X ((√(25−X)) +(√(144−X)) )^2 =13^2 25−X+144−X+2(√((25−X)(144−X))) =169 169−2X+2(√(X^2 −169X+3600)) =169 (√(X^2 −169X+3600)) =X X^2 −169X+3600=X^2 169X=3600 ⇒ 169X=3600 ⇒ x=^3 (√((3600)/(169))) =2,77

$${posons}\:\:{x}^{\mathrm{3}} =\mathrm{X} \\ $$$$\left(\sqrt{\mathrm{25}−\mathrm{X}}\:+\sqrt{\mathrm{144}−\mathrm{X}}\:\right)^{\mathrm{2}} \:=\mathrm{13}^{\mathrm{2}} \\ $$$$\mathrm{25}−\mathrm{X}+\mathrm{144}−\mathrm{X}+\mathrm{2}\sqrt{\left(\mathrm{25}−\mathrm{X}\right)\left(\mathrm{144}−\mathrm{X}\right)}\:=\mathrm{169} \\ $$$$ \\ $$$$\mathrm{169}−\mathrm{2X}+\mathrm{2}\sqrt{\mathrm{X}^{\mathrm{2}} −\mathrm{169X}+\mathrm{3600}}\:=\mathrm{169} \\ $$$$\sqrt{\mathrm{X}^{\mathrm{2}} −\mathrm{169X}+\mathrm{3600}}\:=\mathrm{X} \\ $$$$\mathrm{X}^{\mathrm{2}} −\mathrm{169X}+\mathrm{3600}=\mathrm{X}^{\mathrm{2}} \\ $$$$\mathrm{169X}=\mathrm{3600}\:\:\:\:\Rightarrow \\ $$$$\:\:\mathrm{169X}=\mathrm{3600}\:\:\Rightarrow\:\:\:\:{x}=^{\mathrm{3}} \sqrt{\frac{\mathrm{3600}}{\mathrm{169}}}\:=\mathrm{2},\mathrm{77} \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 02/Oct/22

(√(25−x^3 ))+(√(144−x^3 ))=13 , ∣x∣≤((25))^(1/3) (√(25−x^3 ))=13−(√(144−x^3 )) 25−x^3 =169−26(√(144−x^3 ))+(144−x^3 ) 26(√(144−x^3 ))=288 676(144−x^3 )=82944 ⇒676x^3 =14400 ⇒x^3 =((3600)/(169)) ⇒x=(((3600)/(169)))^(1/3)

$$\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }+\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }=\mathrm{13}\:,\:\mid{x}\mid\leqslant\sqrt[{\mathrm{3}}]{\mathrm{25}} \\ $$$$\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }=\mathrm{13}−\sqrt{\mathrm{144}−{x}^{\mathrm{3}} } \\ $$$$\mathrm{25}−{x}^{\mathrm{3}} =\mathrm{169}−\mathrm{26}\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }+\left(\mathrm{144}−{x}^{\mathrm{3}} \right) \\ $$$$\mathrm{26}\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }=\mathrm{288} \\ $$$$\mathrm{676}\left(\mathrm{144}−{x}^{\mathrm{3}} \right)=\mathrm{82944} \\ $$$$\Rightarrow\mathrm{676}{x}^{\mathrm{3}} =\mathrm{14400} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =\frac{\mathrm{3600}}{\mathrm{169}}\:\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}} \\ $$

Commented by Frix last updated on 02/Oct/22

x_1 =(((3600)/(169)))^(1/3) ; x_2 =ωx_1 ; x_3 =ω^2 x_1 with ω=−(1/2)+((√3)/2)i

$${x}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}};\:{x}_{\mathrm{2}} =\omega{x}_{\mathrm{1}} ;\:{x}_{\mathrm{3}} =\omega^{\mathrm{2}} {x}_{\mathrm{1}} \:\mathrm{with}\:\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$

Commented by Ar Brandon last updated on 02/Oct/22

Thanks for completing

Answered by mr W last updated on 02/Oct/22

let x^3 =t^2 (√(5^2 −t^2 ))+(√(12^2 −t^2 ))=13 t is the altitude of right triangle (1/2)×13×t=(1/2)×5×12 ⇒t=((5×12)/(13))=((60)/(13)) ⇒x=(t^2 )^(1/3) =(((3600)/(169)))^(1/3) ≈2.772

$${let}\:{x}^{\mathrm{3}} ={t}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{5}^{\mathrm{2}} −{t}^{\mathrm{2}} }+\sqrt{\mathrm{12}^{\mathrm{2}} −{t}^{\mathrm{2}} }=\mathrm{13} \\ $$$${t}\:{is}\:{the}\:{altitude}\:{of}\:{right}\:{triangle} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{13}×{t}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}×\mathrm{12} \\ $$$$\Rightarrow{t}=\frac{\mathrm{5}×\mathrm{12}}{\mathrm{13}}=\frac{\mathrm{60}}{\mathrm{13}} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{{t}^{\mathrm{2}} }=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}}\approx\mathrm{2}.\mathrm{772} \\ $$

Commented by mr W last updated on 02/Oct/22

Commented by Rasheed.Sindhi last updated on 02/Oct/22

Deep thinking! It′s only you sir who can see such connectkons!

$$\mathcal{D}{eep}\:{thinking}!\:\mathcal{I}{t}'{s}\:{only}\:{you}\:{sir}\:{who} \\ $$$${can}\:{see}\:{such}\:{connectkons}! \\ $$

Commented by mr W last updated on 02/Oct/22

thanks sir! due to my love for geometry i try at first, if possible, to solve a problem geometrically, sometimes with success, sometimes not.

$${thanks}\:{sir}!\:{due}\:{to}\:{my}\:{love}\:{for}\:{geometry} \\ $$$${i}\:{try}\:{at}\:{first},\:{if}\:{possible},\:{to}\:{solve}\:{a} \\ $$$${problem}\:{geometrically},\:{sometimes} \\ $$$${with}\:{success},\:{sometimes}\:{not}. \\ $$

Commented by Ar Brandon last updated on 02/Oct/22

Sir how can I master geometry too?��

Commented by mr W last updated on 02/Oct/22

i think when you love it, you master it!

$${i}\:{think}\:{when}\:{you}\:{love}\:{it},\:{you}\:{master}\:{it}! \\ $$

Commented by Ar Brandon last updated on 02/Oct/22

Sir, how do we call the type of geometry which you often solve ? For example in Calculus this ∫_0 ^1 ((ln^4 x)/(1+x+x^2 ))dx is integration. But it falls under the advanced part called “special functions”. I feel the geometry you solve is very much advanced than the coordinate geometry most of us saw in secondary school. Does it have a name, Sir?

$$\mathrm{Sir},\:\mathrm{how}\:\mathrm{do}\:\mathrm{we}\:\mathrm{call}\:\mathrm{the}\:\mathrm{type}\:\mathrm{of}\:\mathrm{geometry} \\ $$$$\mathrm{which}\:\mathrm{you}\:\mathrm{often}\:\mathrm{solve}\:? \\ $$$$\mathrm{For}\:\mathrm{example}\:\mathrm{in}\:\mathrm{Calculus}\:\mathrm{this}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{4}} {x}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{dx}\:\mathrm{is}\:\mathrm{integration}. \\ $$$$\mathrm{But}\:\mathrm{it}\:\mathrm{falls}\:\mathrm{under}\:\mathrm{the}\:\mathrm{advanced}\:\mathrm{part}\:\mathrm{called}\:``\mathrm{special}\:\mathrm{functions}''. \\ $$$$\mathrm{I}\:\mathrm{feel}\:\mathrm{the}\:\mathrm{geometry}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{is}\:\mathrm{very}\:\mathrm{much}\:\mathrm{advanced} \\ $$$$\mathrm{than}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{geometry}\:\mathrm{most}\:\mathrm{of}\:\mathrm{us}\:\mathrm{saw}\:\mathrm{in} \\ $$$$\mathrm{secondary}\:\mathrm{school}.\:\mathrm{Does}\:\mathrm{it}\:\mathrm{have}\:\mathrm{a}\:\mathrm{name},\:\mathrm{Sir}? \\ $$

Commented by mr W last updated on 02/Oct/22

i don′t know any “advanced” geometry. what i know is that what people learn in the school: geometry and analytic geometry.

$${i}\:{don}'{t}\:{know}\:{any}\:``{advanced}''\:{geometry}. \\ $$$${what}\:{i}\:{know}\:{is}\:{that}\:{what}\:{people} \\ $$$${learn}\:{in}\:{the}\:{school}:\:{geometry}\:{and} \\ $$$${analytic}\:{geometry}. \\ $$

Commented by Ar Brandon last updated on 02/Oct/22

OK thanks

Answered by Rasheed.Sindhi last updated on 02/Oct/22

(√(25−x^3 )) +(√(144−x^3 ))=13 (√(144−x^3 )) =a , (√(25−x^3 )) =b { ((a+b=13...(i))),((a^2 −b^2 =119)) :} ⇒((a^2 −b^2 )/(a+b))=((119)/(13)) ⇒a−b=((119)/(13))...(ii) (i)+(ii):2a=13+((119)/(13))=((288)/(13)) a=((144)/(13))⇒b=13−((144)/(13))=((25)/(13)) (√(25−x^3 )) =((25)/(13)) 25−x^3 =((625)/(169))⇒x^3 =25−((625)/(169))=((3600)/(169)) x=(((3600)/(169)))^(1/3) , (((3600)/(169)))^(1/3) ω , (((3600)/(169)))^(1/3) ω^2

$$\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:+\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }=\mathrm{13} \\ $$$$\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }\:={a}\:,\:\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:={b} \\ $$$$\begin{cases}{{a}+{b}=\mathrm{13}...\left({i}\right)}\\{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{119}}\end{cases}\:\Rightarrow\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}+{b}}=\frac{\mathrm{119}}{\mathrm{13}} \\ $$$$\Rightarrow{a}−{b}=\frac{\mathrm{119}}{\mathrm{13}}...\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right):\mathrm{2}{a}=\mathrm{13}+\frac{\mathrm{119}}{\mathrm{13}}=\frac{\mathrm{288}}{\mathrm{13}} \\ $$$${a}=\frac{\mathrm{144}}{\mathrm{13}}\Rightarrow{b}=\mathrm{13}−\frac{\mathrm{144}}{\mathrm{13}}=\frac{\mathrm{25}}{\mathrm{13}} \\ $$$$\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:=\frac{\mathrm{25}}{\mathrm{13}} \\ $$$$\mathrm{25}−{x}^{\mathrm{3}} =\frac{\mathrm{625}}{\mathrm{169}}\Rightarrow{x}^{\mathrm{3}} =\mathrm{25}−\frac{\mathrm{625}}{\mathrm{169}}=\frac{\mathrm{3600}}{\mathrm{169}} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}}\:,\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}}\:\omega\:,\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}}\:\omega^{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 02/Oct/22

(√(25−x^3 ))+(√(144−x^3 ))=13...(i) ((√(25−x^3 ))+(√(144−x^3 )) )((√(25−x^3 )) −(√(144−x^3 )) ) =13((√(25−x^3 )) −(√(144−x^3 )) ) 13((√(25−x^3 )) −(√(144−x^3 )) ) =(25−x^3 )−(144−x^3 ) (√(25−x^3 )) −(√(144−x^3 )) =((−119)/(13))...(ii) (i)+(ii): 2(√(25−x^3 )) =13−((119)/(13))=((50)/(13)) 25−x^3 =(((25)/(13)))^2 =((625)/(169)) x^3 =25−((625)/(169))=((3600)/(169)) x=(((3600)/(169)))^(1/3) , (((3600)/(169)))^(1/3) ω,(((3600)/(169)))^(1/3) ω^2

$$\:\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }+\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }=\mathrm{13}...\left({i}\right) \\ $$$$\left(\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }+\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }\:\right)\left(\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:−\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{13}\left(\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:−\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }\:\right) \\ $$$$\mathrm{13}\left(\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:−\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{25}−{x}^{\mathrm{3}} \right)−\left(\mathrm{144}−{x}^{\mathrm{3}} \right) \\ $$$$\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:−\sqrt{\mathrm{144}−{x}^{\mathrm{3}} }\:=\frac{−\mathrm{119}}{\mathrm{13}}...\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}\sqrt{\mathrm{25}−{x}^{\mathrm{3}} }\:=\mathrm{13}−\frac{\mathrm{119}}{\mathrm{13}}=\frac{\mathrm{50}}{\mathrm{13}} \\ $$$$\mathrm{25}−{x}^{\mathrm{3}} \:=\left(\frac{\mathrm{25}}{\mathrm{13}}\right)^{\mathrm{2}} =\frac{\mathrm{625}}{\mathrm{169}} \\ $$$${x}^{\mathrm{3}} =\mathrm{25}−\frac{\mathrm{625}}{\mathrm{169}}=\frac{\mathrm{3600}}{\mathrm{169}} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}}\:,\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}}\:\omega,\sqrt[{\mathrm{3}}]{\frac{\mathrm{3600}}{\mathrm{169}}}\:\omega^{\mathrm{2}} \\ $$

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