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Question Number 155345 by mathdanisur last updated on 29/Sep/21

Solve the equation in R ((5(√(x+1)))/( (√(1 - x + x^2 )) + 2(√(x + 1)))) = 4x^2 - 5x + 5

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{5}\sqrt{\mathrm{x}+\mathrm{1}}}{\:\sqrt{\mathrm{1}\:-\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} }\:+\:\mathrm{2}\sqrt{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\mathrm{4x}^{\mathrm{2}} \:-\:\mathrm{5x}\:+\:\mathrm{5} \\ $$

Commented by MJS_new last updated on 29/Sep/21

no real solution maximum of lhs ≈1.86 minimum of rhs =((55)/(16))

$$\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\mathrm{maximum}\:\mathrm{of}\:\mathrm{lhs}\:\approx\mathrm{1}.\mathrm{86} \\ $$$$\mathrm{minimum}\:\mathrm{of}\:\mathrm{rhs}\:=\frac{\mathrm{55}}{\mathrm{16}} \\ $$

Answered by ArielVyny last updated on 29/Sep/21

((5(√(x+1)))/( (√(1−x+x^2 ))+2(√(x+1))))=4x^2 −5x+5 { ((x≥−1)),((1−x+x^2 #4x+4→x^2 −5x−3#0)) :} ((5(√(x+1))((√(1−x+x^2 ))−2(√(x+1))))/((1−x+x^2 )−(4(x+1))))=4x^2 −5x+5 ((5(√(x+1))((√(1−x+x^2 ))−2(√(x+1))))/(x^2 −5x−3))=4x^2 −5x+5 5(√(x+1))((√(1−x+x^2 )))−10x−10=4x^2 −5x+5 5(√(x+1))((√(1−x+x^2 )))=4x^2 +5x+5 25(x+1)(1−x+x^2 )=(16x^4 +20x^3 +20x^2 )+(20x^3 +25x^2 +25x)+(20x^2 +25x+25) 16x^4 +15x^3 +65x^2 +50x=0 x(16x^3 +15^2 +65x+50)=0 x=0 ou 16x^3 +15x^2 +65x+50=0

$$\frac{\mathrm{5}\sqrt{{x}+\mathrm{1}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }+\mathrm{2}\sqrt{{x}+\mathrm{1}}}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{5} \\ $$$$\begin{cases}{{x}\geqslant−\mathrm{1}}\\{\mathrm{1}−{x}+{x}^{\mathrm{2}} #\mathrm{4}{x}+\mathrm{4}\rightarrow{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}#\mathrm{0}}\end{cases} \\ $$$$\frac{\mathrm{5}\sqrt{{x}+\mathrm{1}}\left(\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }−\mathrm{2}\sqrt{{x}+\mathrm{1}}\right)}{\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)−\left(\mathrm{4}\left({x}+\mathrm{1}\right)\right)}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{5} \\ $$$$\frac{\mathrm{5}\sqrt{{x}+\mathrm{1}}\left(\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }−\mathrm{2}\sqrt{{x}+\mathrm{1}}\right)}{{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{5} \\ $$$$\mathrm{5}\sqrt{{x}+\mathrm{1}}\left(\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }\right)−\mathrm{10}{x}−\mathrm{10}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{5} \\ $$$$\mathrm{5}\sqrt{{x}+\mathrm{1}}\left(\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5} \\ $$$$\mathrm{25}\left({x}+\mathrm{1}\right)\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)=\left(\mathrm{16}{x}^{\mathrm{4}} +\mathrm{20}{x}^{\mathrm{3}} +\mathrm{20}{x}^{\mathrm{2}} \right)+\left(\mathrm{20}{x}^{\mathrm{3}} +\mathrm{25}{x}^{\mathrm{2}} +\mathrm{25}{x}\right)+\left(\mathrm{20}{x}^{\mathrm{2}} +\mathrm{25}{x}+\mathrm{25}\right) \\ $$$$\mathrm{16}{x}^{\mathrm{4}} +\mathrm{15}{x}^{\mathrm{3}} +\mathrm{65}{x}^{\mathrm{2}} +\mathrm{50}{x}=\mathrm{0} \\ $$$${x}\left(\mathrm{16}{x}^{\mathrm{3}} +\mathrm{15}^{\mathrm{2}} +\mathrm{65}{x}+\mathrm{50}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:{ou}\:\mathrm{16}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{65}{x}+\mathrm{50}=\mathrm{0} \\ $$$$ \\ $$

Commented by MJS_new last updated on 29/Sep/21

x=0 is wrong. test it!

$${x}=\mathrm{0}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{test}\:\mathrm{it}! \\ $$

Commented by ArielVyny last updated on 29/Sep/21

yes i have say ′′or′′ to respect ab=0 { ((a=0)),((b=0)) :}or

$${yes}\:{i}\:{have}\:{say}\:''{or}''\:{to}\:{respect}\:{ab}=\mathrm{0}\:\begin{cases}{{a}=\mathrm{0}}\\{{b}=\mathrm{0}}\end{cases}{or} \\ $$

Answered by MJS_new last updated on 29/Sep/21

((5(√(x+1)))/( (√(x^2 −x+1))+2(√(x+1))))=4x^2 −5x+5 −(8x^2 −10x+5)(√(x+1))=(4x^2 −5x+5)(√(x^2 −x+1)) squaring (might introduce false solutions) and teansforming x^2 (x^4 −((15)/2)x^3 +((217)/(16))x^2 −((175)/(16))x+((15)/4))=0 obviously x=0 is false x^4 −((15)/2)x^3 +((217)/(16))x^2 −((175)/(16))x+((15)/4)=0 (x^2 −((25)/4)x+5)(x^2 −(5/4)x+(3/4))=0 x=((25)/8)±((√(305))/8) [both false] x=(5/8)±((√(23))/8)i [both true]

$$\frac{\mathrm{5}\sqrt{{x}+\mathrm{1}}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}+\mathrm{2}\sqrt{{x}+\mathrm{1}}}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{5} \\ $$$$−\left(\mathrm{8}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{5}\right)\sqrt{{x}+\mathrm{1}}=\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{5}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\mathrm{squaring}\:\left(\mathrm{might}\:\mathrm{introduce}\:\mathrm{false}\:\mathrm{solutions}\right) \\ $$$$\mathrm{and}\:\mathrm{teansforming} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{4}} −\frac{\mathrm{15}}{\mathrm{2}}{x}^{\mathrm{3}} +\frac{\mathrm{217}}{\mathrm{16}}{x}^{\mathrm{2}} −\frac{\mathrm{175}}{\mathrm{16}}{x}+\frac{\mathrm{15}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\mathrm{obviously}\:{x}=\mathrm{0}\:\mathrm{is}\:\mathrm{false} \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{15}}{\mathrm{2}}{x}^{\mathrm{3}} +\frac{\mathrm{217}}{\mathrm{16}}{x}^{\mathrm{2}} −\frac{\mathrm{175}}{\mathrm{16}}{x}+\frac{\mathrm{15}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{4}}{x}+\mathrm{5}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$${x}=\frac{\mathrm{25}}{\mathrm{8}}\pm\frac{\sqrt{\mathrm{305}}}{\mathrm{8}}\:\left[\mathrm{both}\:\mathrm{false}\right] \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{8}}\pm\frac{\sqrt{\mathrm{23}}}{\mathrm{8}}\mathrm{i}\:\left[\mathrm{both}\:\mathrm{true}\right] \\ $$

Commented by mathdanisur last updated on 29/Sep/21

Very nice solution Ser thank you

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{solution}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mathdanisur last updated on 30/Sep/21

But dear Ser, we solve in R

$$\mathrm{But}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{we}\:\mathrm{solve}\:\mathrm{in}\:\mathbb{R} \\ $$

Commented by MJS_new last updated on 30/Sep/21

yes. as I said, no solution in R

$$\mathrm{yes}.\:\mathrm{as}\:\mathrm{I}\:\mathrm{said},\:\mathrm{no}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{R} \\ $$

Commented by mathdanisur last updated on 30/Sep/21

THANKYOU DEAR SER

$$\mathrm{THANKYOU}\:\mathrm{DEAR}\:\mathrm{SER} \\ $$

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