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Question Number 128166 by bramlexs22 last updated on 05/Jan/21

 (√(x−2)) + (√(x+3)) +(√(4x+1)) = 10   for x∈R

$$\:\sqrt{\mathrm{x}−\mathrm{2}}\:+\:\sqrt{\mathrm{x}+\mathrm{3}}\:+\sqrt{\mathrm{4x}+\mathrm{1}}\:=\:\mathrm{10} \\ $$$$\:\mathrm{for}\:\mathrm{x}\in\mathbb{R}\: \\ $$

Answered by liberty last updated on 05/Jan/21

 (√(4x+1)) = [ −(√(x+3))−(√(x−2))+10 ]   4x+1 = 2(√(x−2 )) (√(x+3)) −20(√(x+3)) +2x −20(√(x−2)) +101   (√(x+3)) = ((x+10(√(x−2))−50)/( (√(x−2))−10))   x+3 = (x^2 /(x−20(√(x−2))+98)) + ((20x(√(x−2)))/(x−20(√(x−2)) +98)) −((1000(√(x−2)))/(x−20(√(x−2))+98)) +((2300)/(x−20(√(x−2))+98))    (√(x−2)) =((101x−2006)/(40x−940))   x−2 = (((101x−2006)^2 )/((40x−940)^2 ))   x = −((399(√(401))−79001)/(3200)) ; x=((399(√(401)) +79001)/(3200)) ; x=6

$$\:\sqrt{\mathrm{4x}+\mathrm{1}}\:=\:\left[\:−\sqrt{\mathrm{x}+\mathrm{3}}−\sqrt{\mathrm{x}−\mathrm{2}}+\mathrm{10}\:\right] \\ $$$$\:\mathrm{4x}+\mathrm{1}\:=\:\mathrm{2}\sqrt{\mathrm{x}−\mathrm{2}\:}\:\sqrt{\mathrm{x}+\mathrm{3}}\:−\mathrm{20}\sqrt{\mathrm{x}+\mathrm{3}}\:+\mathrm{2x}\:−\mathrm{20}\sqrt{\mathrm{x}−\mathrm{2}}\:+\mathrm{101} \\ $$$$\:\sqrt{\mathrm{x}+\mathrm{3}}\:=\:\frac{\mathrm{x}+\mathrm{10}\sqrt{\mathrm{x}−\mathrm{2}}−\mathrm{50}}{\:\sqrt{\mathrm{x}−\mathrm{2}}−\mathrm{10}} \\ $$$$\:\mathrm{x}+\mathrm{3}\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{20}\sqrt{\mathrm{x}−\mathrm{2}}+\mathrm{98}}\:+\:\frac{\mathrm{20x}\sqrt{\mathrm{x}−\mathrm{2}}}{\mathrm{x}−\mathrm{20}\sqrt{\mathrm{x}−\mathrm{2}}\:+\mathrm{98}}\:−\frac{\mathrm{1000}\sqrt{\mathrm{x}−\mathrm{2}}}{\mathrm{x}−\mathrm{20}\sqrt{\mathrm{x}−\mathrm{2}}+\mathrm{98}}\:+\frac{\mathrm{2300}}{\mathrm{x}−\mathrm{20}\sqrt{\mathrm{x}−\mathrm{2}}+\mathrm{98}}\: \\ $$$$\:\sqrt{\mathrm{x}−\mathrm{2}}\:=\frac{\mathrm{101x}−\mathrm{2006}}{\mathrm{40x}−\mathrm{940}} \\ $$$$\:\mathrm{x}−\mathrm{2}\:=\:\frac{\left(\mathrm{101x}−\mathrm{2006}\right)^{\mathrm{2}} }{\left(\mathrm{40x}−\mathrm{940}\right)^{\mathrm{2}} } \\ $$$$\:\mathrm{x}\:=\:−\frac{\mathrm{399}\sqrt{\mathrm{401}}−\mathrm{79001}}{\mathrm{3200}}\:;\:\mathrm{x}=\frac{\mathrm{399}\sqrt{\mathrm{401}}\:+\mathrm{79001}}{\mathrm{3200}}\:;\:\mathrm{x}=\mathrm{6}\: \\ $$$$ \\ $$

Commented by MJS_new last updated on 05/Jan/21

obviously x=6 is the only real solution.  by squaring you introduce false solutions,  you have to check each single one inserting  it in the original equation

$$\mathrm{obviously}\:{x}=\mathrm{6}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{real}\:\mathrm{solution}. \\ $$$$\mathrm{by}\:\mathrm{squaring}\:\mathrm{you}\:\mathrm{introduce}\:\mathrm{false}\:\mathrm{solutions}, \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{check}\:\mathrm{each}\:\mathrm{single}\:\mathrm{one}\:\mathrm{inserting} \\ $$$$\mathrm{it}\:\mathrm{in}\:\mathrm{the}\:\mathrm{original}\:\mathrm{equation} \\ $$

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