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Question Number 196119 by mr W last updated on 18/Aug/23
solve (√(100−x^2 ))+(√(64−x^2 ))=12
$${solve} \\ $$$$\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }=\mathrm{12} \\ $$
Answered by cortano12 last updated on 18/Aug/23
(√((10−x)(10+x))) +(√((8−x)(8+x))) = 12 { ((8−x=u)),((8+x=v)) :} (√((u+2)(v+2))) + (√(u.v)) = 12 (√(uv+2(u+v)+4)) = 12−(√(uv)) (√(uv+36)) = 12−(√(uv)) uv + 36 = 144−24(√(uv)) + uv 2(√(uv)) = 12−3=9 uv= ((81)/4) ; 64−x^2 = ((81)/4) 256−4x^2 = 81 175−4x^2 = 0 (5(√7)−2x)(5(√7)+2x)=0 x = ± ((5(√7))/2)
$$\:\sqrt{\left(\mathrm{10}−\mathrm{x}\right)\left(\mathrm{10}+\mathrm{x}\right)}\:+\sqrt{\left(\mathrm{8}−\mathrm{x}\right)\left(\mathrm{8}+\mathrm{x}\right)}\:=\:\mathrm{12} \\ $$$$\:\begin{cases}{\mathrm{8}−\mathrm{x}=\mathrm{u}}\\{\mathrm{8}+\mathrm{x}=\mathrm{v}}\end{cases} \\ $$$$\:\sqrt{\left(\mathrm{u}+\mathrm{2}\right)\left(\mathrm{v}+\mathrm{2}\right)}\:+\:\sqrt{\mathrm{u}.\mathrm{v}}\:=\:\mathrm{12} \\ $$$$\:\:\sqrt{\mathrm{uv}+\mathrm{2}\left(\mathrm{u}+\mathrm{v}\right)+\mathrm{4}}\:=\:\mathrm{12}−\sqrt{\mathrm{uv}} \\ $$$$\:\:\sqrt{\mathrm{uv}+\mathrm{36}}\:=\:\mathrm{12}−\sqrt{\mathrm{uv}} \\ $$$$\:\:\mathrm{uv}\:+\:\mathrm{36}\:=\:\mathrm{144}−\mathrm{24}\sqrt{\mathrm{uv}}\:+\:\mathrm{uv} \\ $$$$\:\:\:\mathrm{2}\sqrt{\mathrm{uv}}\:=\:\mathrm{12}−\mathrm{3}=\mathrm{9} \\ $$$$\:\:\:\mathrm{uv}=\:\frac{\mathrm{81}}{\mathrm{4}}\:;\:\mathrm{64}−\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{81}}{\mathrm{4}} \\ $$$$\:\:\:\mathrm{256}−\mathrm{4x}^{\mathrm{2}} \:=\:\mathrm{81}\: \\ $$$$\:\:\:\mathrm{175}−\mathrm{4x}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\:\:\left(\mathrm{5}\sqrt{\mathrm{7}}−\mathrm{2x}\right)\left(\mathrm{5}\sqrt{\mathrm{7}}+\mathrm{2x}\right)=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\pm\:\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}}\: \\ $$$$\: \\ $$
Commented by mr W last updated on 18/Aug/23
thanks!
$${thanks}! \\ $$
Answered by Rasheed.Sindhi last updated on 18/Aug/23
(√(100−x^2 ))+(√(64−x^2 ))=12....(i) ((√(100−x^2 ))+(√(64−x^2 )) )((√(100−x^2 )) −(√(64−x^2 )) )=12((√(100−x^2 )) −(√(64−x^2 )) ) ((√(100−x^2 )) )^2 +((√(64−x^2 )) )^2 =12((√(100−x^2 )) −(√(64−x^2 )) ) (√(100−x^2 )) −(√(64−x^2 )) =((36)/(12))=3....(ii) (i)+(ii): 2(√(100−x^2 )) =15 100−x^2 =((225)/4) x^2 =100−((225)/4)=((175)/4) x=((±5(√7))/2)
$$\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }=\mathrm{12}….\left({i}\right) \\ $$$$\left(\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:\right)\left(\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:−\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:\right)=\mathrm{12}\left(\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:−\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:\right) \\ $$$$\left(\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:\right)^{\mathrm{2}} =\mathrm{12}\left(\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:−\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:\right) \\ $$$$\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:−\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:=\frac{\mathrm{36}}{\mathrm{12}}=\mathrm{3}….\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:=\mathrm{15} \\ $$$$\mathrm{100}−{x}^{\mathrm{2}} =\frac{\mathrm{225}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} =\mathrm{100}−\frac{\mathrm{225}}{\mathrm{4}}=\frac{\mathrm{175}}{\mathrm{4}} \\ $$$${x}=\frac{\pm\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 18/Aug/23
thanks!
$${thanks}! \\ $$
Answered by gatocomcirrose last updated on 18/Aug/23
u=100−x^2 v=64−x^2 u−v=36⇒((√u)+(√v))((√u)−(√v))=36 (√u)+(√v)=12 ⇒(√u)−(√v)=3 ⇒(√u)=((15)/2)⇒u=((225)/4)⇒v=((81)/4) ⇒100−x^2 =((225)/4)⇒x=±((5(√7))/2)
$$\mathrm{u}=\mathrm{100}−\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{v}=\mathrm{64}−\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{u}−\mathrm{v}=\mathrm{36}\Rightarrow\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)\left(\sqrt{\mathrm{u}}−\sqrt{\mathrm{v}}\right)=\mathrm{36} \\ $$$$\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}=\mathrm{12} \\ $$$$\Rightarrow\sqrt{\mathrm{u}}−\sqrt{\mathrm{v}}=\mathrm{3} \\ $$$$\Rightarrow\sqrt{\mathrm{u}}=\frac{\mathrm{15}}{\mathrm{2}}\Rightarrow\mathrm{u}=\frac{\mathrm{225}}{\mathrm{4}}\Rightarrow\mathrm{v}=\frac{\mathrm{81}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{100}−\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{225}}{\mathrm{4}}\Rightarrow\mathrm{x}=\pm\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 18/Aug/23
thanks!
$${thanks}! \\ $$
Answered by mr W last updated on 18/Aug/23
Commented by mr W last updated on 18/Aug/23
cos α=((12^2 +10^2 −8^2 )/(2×12×10))=(3/4) ⇒sin α=((√7)/4) ⇒x=10×((√7)/4)=((5(√7))/2) ✓ x=−((5(√7))/2) is also a root.
$$\mathrm{cos}\:\alpha=\frac{\mathrm{12}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{12}×\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{10}×\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}=\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\checkmark \\ $$$$\:{x}=−\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}}\:{is}\:{also}\:{a}\:{root}. \\ $$
Commented by Rasheed.Sindhi last updated on 18/Aug/23
A novel idea!
$${A}\:{novel}\:{idea}! \\ $$
Answered by Frix last updated on 19/Aug/23
(√(100−x^2 ))+(√(64−y^2 ))=12 Let x=10sin α ∧y=8sin β ⇒ 5cos α +4cos β =6 ⇒ cos β =((6−5cos α)/4) But x=y ⇒ 5sin α =4sin β ⇒ sin β =((5sin α)/4) cos^2 β +sin^2 β =1 ((36−60cos α +25cos^2 α)/(16))+((25sin^2 α)/(16))=1 ⇒ cos α =(3/4) ⇒ sin α =±((√7)/4) ⇒ x=±((5(√7))/2)
$$\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{64}−{y}^{\mathrm{2}} }=\mathrm{12} \\ $$$$\mathrm{Let}\:{x}=\mathrm{10sin}\:\alpha\:\wedge{y}=\mathrm{8sin}\:\beta\:\Rightarrow \\ $$$$\mathrm{5cos}\:\alpha\:+\mathrm{4cos}\:\beta\:=\mathrm{6}\:\Rightarrow\:\mathrm{cos}\:\beta\:=\frac{\mathrm{6}−\mathrm{5cos}\:\alpha}{\mathrm{4}} \\ $$$$\mathrm{But}\:{x}={y}\:\Rightarrow \\ $$$$\mathrm{5sin}\:\alpha\:=\mathrm{4sin}\:\beta\:\Rightarrow\:\mathrm{sin}\:\beta\:=\frac{\mathrm{5sin}\:\alpha}{\mathrm{4}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\beta\:+\mathrm{sin}^{\mathrm{2}} \:\beta\:=\mathrm{1} \\ $$$$\frac{\mathrm{36}−\mathrm{60cos}\:\alpha\:+\mathrm{25cos}^{\mathrm{2}} \:\alpha}{\mathrm{16}}+\frac{\mathrm{25sin}^{\mathrm{2}} \:\alpha}{\mathrm{16}}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}\:\alpha\:=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow\:\mathrm{sin}\:\alpha\:=\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}\:\Rightarrow\:{x}=\pm\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 19/Aug/23
∩i⊂∈!
$$\cap\boldsymbol{\mathrm{i}}\subset\in! \\ $$
Commented by Frix last updated on 19/Aug/23
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