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Question Number 171043 by mathlove last updated on 06/Jun/22

A∈R  A=(((√(x−2))+x+3)/( (√(4−2x))+x−1))          faind A=?

$${A}\in{R} \\ $$$${A}=\frac{\sqrt{{x}−\mathrm{2}}+{x}+\mathrm{3}}{\:\sqrt{\mathrm{4}−\mathrm{2}{x}}+{x}−\mathrm{1}}\:\:\:\:\:\:\:\:\:\:{faind}\:{A}=? \\ $$

Answered by MJS_new last updated on 07/Jun/22

(√(x−2))∈R ⇒ x≥2  (√(4−2x))=(√2)(√(2−x))∈R ⇒ x≤2  ⇒ x=2  ⇒ A=5

$$\sqrt{{x}−\mathrm{2}}\in\mathbb{R}\:\Rightarrow\:{x}\geqslant\mathrm{2} \\ $$$$\sqrt{\mathrm{4}−\mathrm{2}{x}}=\sqrt{\mathrm{2}}\sqrt{\mathrm{2}−{x}}\in\mathbb{R}\:\Rightarrow\:{x}\leqslant\mathrm{2} \\ $$$$\Rightarrow\:{x}=\mathrm{2} \\ $$$$\Rightarrow\:{A}=\mathrm{5} \\ $$

Commented by Rasheed.Sindhi last updated on 07/Jun/22

Great sir!

$$\mathbb{G}\boldsymbol{\mathrm{reat}}\:\boldsymbol{\mathrm{sir}}! \\ $$

Commented by MJS_new last updated on 07/Jun/22

there are solutions with x∉R∧A∈R  simply set A∈R and solve for x∈C  A=5 ⇒ x=2∨x=−((17)/(16))±((5(√2))/8)i

$$\mathrm{there}\:\mathrm{are}\:\mathrm{solutions}\:\mathrm{with}\:{x}\notin\mathbb{R}\wedge{A}\in\mathbb{R} \\ $$$$\mathrm{simply}\:\mathrm{set}\:{A}\in\mathbb{R}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$${A}=\mathrm{5}\:\Rightarrow\:{x}=\mathrm{2}\vee{x}=−\frac{\mathrm{17}}{\mathrm{16}}\pm\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{i} \\ $$

Commented by mathlove last updated on 08/Jun/22

thanks

$${thanks} \\ $$

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