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Question Number 88873 by jagoll last updated on 13/Apr/20

3+(√(x^2 −5)) > ∣x−1∣

$$\mathrm{3}+\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\:>\:\mid{x}−\mathrm{1}\mid\: \\ $$

Commented byjohn santu last updated on 13/Apr/20

⇒(3+(√(x^2 −5)))^2  > (x−1)^2   6(√(x^2 −5)) +4+x^2  > x^2 −2x+1  6(√(x^2 −5)) > −2x−3   (i) −2x−3 < 0 ∧ x^2 −5 ≥ 0  ⇒x > −(3/2) ∧ x≤−(√5) ∨ x≥ (√5)  we get x ≥ (√5)   (ii)−2x−3 > 0 ⇒36(x^2 −5) >  4x^2 +12x+9 ⇒x < −(9/4)  the solution : x <−(9/4) ∨ x ≥(√5)

$$\Rightarrow\left(\mathrm{3}+\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\right)^{\mathrm{2}} \:>\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$ $$\mathrm{6}\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\:+\mathrm{4}+{x}^{\mathrm{2}} \:>\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1} \\ $$ $$\mathrm{6}\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\:>\:−\mathrm{2}{x}−\mathrm{3}\: \\ $$ $$\left({i}\right)\:−\mathrm{2}{x}−\mathrm{3}\:<\:\mathrm{0}\:\wedge\:{x}^{\mathrm{2}} −\mathrm{5}\:\geqslant\:\mathrm{0} \\ $$ $$\Rightarrow{x}\:>\:−\frac{\mathrm{3}}{\mathrm{2}}\:\wedge\:{x}\leqslant−\sqrt{\mathrm{5}}\:\vee\:{x}\geqslant\:\sqrt{\mathrm{5}} \\ $$ $${we}\:{get}\:{x}\:\geqslant\:\sqrt{\mathrm{5}}\: \\ $$ $$\left({ii}\right)−\mathrm{2}{x}−\mathrm{3}\:>\:\mathrm{0}\:\Rightarrow\mathrm{36}\left({x}^{\mathrm{2}} −\mathrm{5}\right)\:> \\ $$ $$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{9}\:\Rightarrow{x}\:<\:−\frac{\mathrm{9}}{\mathrm{4}} \\ $$ $${the}\:{solution}\::\:{x}\:<−\frac{\mathrm{9}}{\mathrm{4}}\:\vee\:{x}\:\geqslant\sqrt{\mathrm{5}}\: \\ $$

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