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2-x-3-x-3-x-4-x-2-x-4-x-x-2-

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Question Number 159587 by cortano last updated on 19/Nov/21
(√(2−x)) (√(3−x)) + (√(3−x)) (√(4−x)) + (√(2−x)) (√(4−x)) = x+2
$$\:\sqrt{\mathrm{2}−{x}}\:\sqrt{\mathrm{3}−{x}}\:+\:\sqrt{\mathrm{3}−{x}}\:\sqrt{\mathrm{4}−{x}}\:+\:\sqrt{\mathrm{2}−{x}}\:\sqrt{\mathrm{4}−{x}}\:=\:{x}+\mathrm{2} \\ $$$$ \\ $$
Commented by mr W last updated on 19/Nov/21
we can use my method in Q159527 for solving this kind of equations.
$${we}\:{can}\:{use}\:{my}\:{method}\:{in}\:{Q}\mathrm{159527} \\ $$$${for}\:{solving}\:{this}\:{kind}\:{of}\:{equations}. \\ $$
Answered by mr W last updated on 19/Nov/21
((√(2−x))+(√(3−x))+(√(4−x)))^2 =2−x+3−x+4−x+2(...LHS...) ((√(2−x))+(√(3−x))+(√(4−x)))^2 =9−3x+2(x+2) ((√(2−x))+(√(3−x))+(√(4−x)))^2 =13−x (√(2−x))+(√(3−x))+(√(4−x))=(√(13−x)) (√(2−x))+(√(3−x))=(√(13−x))−(√(4−x)) ((√(2−x))+(√(3−x)))^2 =((√(13−x))−(√(4−x)))^2 2−x+3−x+2(√((2−x)(3−x)))=13−x+4−x−2(√((13−x)(4−x))) (√((2−x)(3−x)))=6−(√((13−x)(4−x))) (2−x)(3−x)=36+(13−x)(4−x)−12(√((13−x)(4−x))) 6−5x+x^2 =36+52−17x+x^2 −12(√((13−x)(4−x))) 41−6x=6(√((13−x)(4−x))) (41−6x)^2 =36(13−x)(4−x) 120x=191 ⇒x=((191)/(120))
$$\left(\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}+\sqrt{\mathrm{4}−{x}}\right)^{\mathrm{2}} =\mathrm{2}−{x}+\mathrm{3}−{x}+\mathrm{4}−{x}+\mathrm{2}\left(…{LHS}…\right) \\ $$$$\left(\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}+\sqrt{\mathrm{4}−{x}}\right)^{\mathrm{2}} =\mathrm{9}−\mathrm{3}{x}+\mathrm{2}\left({x}+\mathrm{2}\right) \\ $$$$\left(\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}+\sqrt{\mathrm{4}−{x}}\right)^{\mathrm{2}} =\mathrm{13}−{x} \\ $$$$\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}+\sqrt{\mathrm{4}−{x}}=\sqrt{\mathrm{13}−{x}} \\ $$$$\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}=\sqrt{\mathrm{13}−{x}}−\sqrt{\mathrm{4}−{x}} \\ $$$$\left(\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{3}−{x}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{13}−{x}}−\sqrt{\mathrm{4}−{x}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}−{x}+\mathrm{3}−{x}+\mathrm{2}\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}=\mathrm{13}−{x}+\mathrm{4}−{x}−\mathrm{2}\sqrt{\left(\mathrm{13}−{x}\right)\left(\mathrm{4}−{x}\right)} \\ $$$$\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}=\mathrm{6}−\sqrt{\left(\mathrm{13}−{x}\right)\left(\mathrm{4}−{x}\right)} \\ $$$$\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)=\mathrm{36}+\left(\mathrm{13}−{x}\right)\left(\mathrm{4}−{x}\right)−\mathrm{12}\sqrt{\left(\mathrm{13}−{x}\right)\left(\mathrm{4}−{x}\right)} \\ $$$$\mathrm{6}−\mathrm{5}{x}+{x}^{\mathrm{2}} =\mathrm{36}+\mathrm{52}−\mathrm{17}{x}+{x}^{\mathrm{2}} −\mathrm{12}\sqrt{\left(\mathrm{13}−{x}\right)\left(\mathrm{4}−{x}\right)} \\ $$$$\mathrm{41}−\mathrm{6}{x}=\mathrm{6}\sqrt{\left(\mathrm{13}−{x}\right)\left(\mathrm{4}−{x}\right)} \\ $$$$\left(\mathrm{41}−\mathrm{6}{x}\right)^{\mathrm{2}} =\mathrm{36}\left(\mathrm{13}−{x}\right)\left(\mathrm{4}−{x}\right) \\ $$$$\mathrm{120}{x}=\mathrm{191} \\ $$$$\Rightarrow{x}=\frac{\mathrm{191}}{\mathrm{120}} \\ $$
Commented by cortano last updated on 19/Nov/21
nice
$${nice} \\ $$

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