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2x-2-5x-2-x-2-x-2-x-2-3x-2-

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Question Number 135493 by EDWIN88 last updated on 13/Mar/21
(√(2x^2 −5x+2)) −(√(x^2 −x−2)) = (√(x^2 −3x+2))
$$\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}\:=\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$
Answered by john_santu last updated on 13/Mar/21
(1) { ((2x^2 −5x+2≥0)),((x^2 −x−2≥0)),((x^2 −3x+2≥0)) :} { ((x≥2)),((x≤−1)) :} consider (√(f(x).g(x))) = (√(∣f(x)∣.∣g(x)∣)) ; f(x),g(x)≥0 (2)(√((2x−1)(x−2)))−(√((x−2)(x+1)))−(√((x−2)(x−1))) = 0 ⇒(√(∣x−2∣)) {(√(∣2x−1∣)) −(√(∣x+1∣)) −(√(∣x−1∣)) }= 0 (3) x = 2 (4) for x≤−1 ⇒(√(1−2x))−(√(−x−1))−(√(1−x)) = 0 ⇒ 1−2x = −2x+2(√((−x−1)(1−x))) ⇒ 1 = 4(x^2 −1) ⇒ x=−((√5)/2) (5) for x≥ 2⇒(√(2x−1)) = (√(x+1)) +(√(x−1)) ⇒2x−1 = 2x+2(√(x^2 −1)) ⇒−1 = 2(√(x^2 −1)) ( has no solution) Therefore the solution is {−((√5)/2); 2 }
$$\left(\mathrm{1}\right)\begin{cases}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}\geqslant\mathrm{0}}\\{{x}^{\mathrm{2}} −{x}−\mathrm{2}\geqslant\mathrm{0}}\\{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}\geqslant\mathrm{0}}\end{cases}\begin{cases}{{x}\geqslant\mathrm{2}}\\{{x}\leqslant−\mathrm{1}}\end{cases} \\ $$$${consider}\:\sqrt{{f}\left({x}\right).{g}\left({x}\right)}\:=\:\sqrt{\mid{f}\left({x}\right)\mid.\mid{g}\left({x}\right)\mid}\:;\:{f}\left({x}\right),{g}\left({x}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\sqrt{\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}−\sqrt{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}−\sqrt{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}\:=\:\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mid{x}−\mathrm{2}\mid}\:\left\{\sqrt{\mid\mathrm{2}{x}−\mathrm{1}\mid}\:−\sqrt{\mid{x}+\mathrm{1}\mid}\:−\sqrt{\mid{x}−\mathrm{1}\mid}\:\right\}=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:{x}\:=\:\mathrm{2} \\ $$$$\left(\mathrm{4}\right)\:{for}\:{x}\leqslant−\mathrm{1}\:\Rightarrow\sqrt{\mathrm{1}−\mathrm{2}{x}}−\sqrt{−{x}−\mathrm{1}}−\sqrt{\mathrm{1}−{x}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{1}−\mathrm{2}{x}\:=\:−\mathrm{2}{x}+\mathrm{2}\sqrt{\left(−{x}−\mathrm{1}\right)\left(\mathrm{1}−{x}\right)} \\ $$$$\Rightarrow\:\mathrm{1}\:=\:\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:\Rightarrow\:{x}=−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\mathrm{5}\right)\:{for}\:{x}\geqslant\:\mathrm{2}\Rightarrow\sqrt{\mathrm{2}{x}−\mathrm{1}}\:=\:\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}{x}−\mathrm{1}\:=\:\mathrm{2}{x}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow−\mathrm{1}\:=\:\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\left(\:{has}\:{no}\:{solution}\right) \\ $$$${Therefore}\:{the}\:{solution}\:{is}\:\left\{−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}};\:\mathrm{2}\:\right\} \\ $$

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