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Question Number 137415 by mr W last updated on 02/Apr/21
solve  x+(√(x(x+1)))+(√(x(x+2)))+(√((x+1)(x+2)))=2
$${solve} \\ $$$${x}+\sqrt{{x}\left({x}+\mathrm{1}\right)}+\sqrt{{x}\left({x}+\mathrm{2}\right)}+\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}=\mathrm{2} \\ $$
Commented by MJS_new last updated on 03/Apr/21
x_1 =(1/(24))  x_2 ≈−2.73908312241  x_2  is a solution of x^4 +x^3 −4x^2 +2x−(1/4)=0  which I couldn′t solve exactly and there are  no complex solutions.
$${x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${x}_{\mathrm{2}} \approx−\mathrm{2}.\mathrm{73908312241} \\ $$$${x}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{of}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{solve}\:\mathrm{exactly}\:\mathrm{and}\:\mathrm{there}\:\mathrm{are} \\ $$$$\mathrm{no}\:\mathrm{complex}\:\mathrm{solutions}. \\ $$
Commented by liberty last updated on 03/Apr/21
how to get x^4 +x^3 −4x^2 +2x−(1/4)=0?
$${how}\:{to}\:{get}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0}? \\ $$
Commented by MJS_new last updated on 03/Apr/21
(√A)+(√B)=D−(√C)  squaring, transforming  2((√(AB))+D(√C))=C+D^2 −A−B  squaring, transforming  8D(√A)(√B)(√C)=A^2 +B^2 +C^2 +D^4 −2(AB+AC+AD^2 +BC+BD^2 +CD^2 )  squaring, transforming, inserting  x^5 +((23)/(24))x^4 −((97)/(24))x^3 +((13)/6)x^2 −(1/3)x+(1/(96))=0  trying factors of (1/(96))  (x−(1/(24)))(x^4 +x^3 −4x^2 +2x−(1/4))=0  we can find 2 square factors  let x=z−(1/4)  z^4 −((35)/8)z^2 +((33)/8)z−((259)/(256))=0  (z^2 −az−b)(z^2 +az−c)=0  ⇒  (1)  a^2 +b+c−((35)/8)=0  (2)  ab−ac+((33)/8)=0  (3)  bc+((259)/(256))=0  (1)  c=−a^2 −b+((35)/8)  (2)  b=−(a^2 /2)−((33)/(16a))+((35)/(16)) ⇒ c=−(a^2 /2)+((33)/(16a))+((35)/(16))  (3)  ⇒ a^6 −((35)/4)a^4 +((371)/(16))a^2 −((1089)/(64))=0  let a=(√(r+((35)/(12))))  r^3 −(7/3)r+((107)/(108))=0  sadly this has no useable solution...  r_1 =((2(√7))/3)sin ((arcsin ((107(√7))/(392)))/3)  r_2 =((2(√7))/3)cos ((π+2arcsin ((107(√7))/(392)))/6)  r_3 =−((2(√7))/3)sin ((π+arcsin ((107(√7))/(392)))/3)  ⇒ this makes no sense, it′s “cheaper” to  approximately solve  x^4 +x^3 −4x^2 +2x−(1/4)=0  and check the solutions (squaring causes  false solutions)  x_1 ≈−2.73908 ★  x_2 ≈.200728 wrong  x_3 ≈.399135 wrong  x_4 ≈1.13922 wrong
$$\sqrt{{A}}+\sqrt{{B}}={D}−\sqrt{{C}} \\ $$$$\mathrm{squaring},\:\mathrm{transforming} \\ $$$$\mathrm{2}\left(\sqrt{{AB}}+{D}\sqrt{{C}}\right)={C}+{D}^{\mathrm{2}} −{A}−{B} \\ $$$$\mathrm{squaring},\:\mathrm{transforming} \\ $$$$\mathrm{8}{D}\sqrt{{A}}\sqrt{{B}}\sqrt{{C}}={A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{C}^{\mathrm{2}} +{D}^{\mathrm{4}} −\mathrm{2}\left({AB}+{AC}+{AD}^{\mathrm{2}} +{BC}+{BD}^{\mathrm{2}} +{CD}^{\mathrm{2}} \right) \\ $$$$\mathrm{squaring},\:\mathrm{transforming},\:\mathrm{inserting} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{23}}{\mathrm{24}}{x}^{\mathrm{4}} −\frac{\mathrm{97}}{\mathrm{24}}{x}^{\mathrm{3}} +\frac{\mathrm{13}}{\mathrm{6}}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{1}}{\mathrm{96}}=\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:\frac{\mathrm{1}}{\mathrm{96}} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{24}}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors} \\ $$$$\mathrm{let}\:{x}={z}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${z}^{\mathrm{4}} −\frac{\mathrm{35}}{\mathrm{8}}{z}^{\mathrm{2}} +\frac{\mathrm{33}}{\mathrm{8}}{z}−\frac{\mathrm{259}}{\mathrm{256}}=\mathrm{0} \\ $$$$\left({z}^{\mathrm{2}} −{az}−{b}\right)\left({z}^{\mathrm{2}} +{az}−{c}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\:{a}^{\mathrm{2}} +{b}+{c}−\frac{\mathrm{35}}{\mathrm{8}}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:{ab}−{ac}+\frac{\mathrm{33}}{\mathrm{8}}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:{bc}+\frac{\mathrm{259}}{\mathrm{256}}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\:{c}=−{a}^{\mathrm{2}} −{b}+\frac{\mathrm{35}}{\mathrm{8}} \\ $$$$\left(\mathrm{2}\right)\:\:{b}=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{33}}{\mathrm{16}{a}}+\frac{\mathrm{35}}{\mathrm{16}}\:\Rightarrow\:{c}=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{33}}{\mathrm{16}{a}}+\frac{\mathrm{35}}{\mathrm{16}} \\ $$$$\left(\mathrm{3}\right)\:\:\Rightarrow\:{a}^{\mathrm{6}} −\frac{\mathrm{35}}{\mathrm{4}}{a}^{\mathrm{4}} +\frac{\mathrm{371}}{\mathrm{16}}{a}^{\mathrm{2}} −\frac{\mathrm{1089}}{\mathrm{64}}=\mathrm{0} \\ $$$$\mathrm{let}\:{a}=\sqrt{{r}+\frac{\mathrm{35}}{\mathrm{12}}} \\ $$$${r}^{\mathrm{3}} −\frac{\mathrm{7}}{\mathrm{3}}{r}+\frac{\mathrm{107}}{\mathrm{108}}=\mathrm{0} \\ $$$$\mathrm{sadly}\:\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:\mathrm{useable}\:\mathrm{solution}… \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{arcsin}\:\frac{\mathrm{107}\sqrt{\mathrm{7}}}{\mathrm{392}}}{\mathrm{3}} \\ $$$${r}_{\mathrm{2}} =\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi+\mathrm{2arcsin}\:\frac{\mathrm{107}\sqrt{\mathrm{7}}}{\mathrm{392}}}{\mathrm{6}} \\ $$$${r}_{\mathrm{3}} =−\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi+\mathrm{arcsin}\:\frac{\mathrm{107}\sqrt{\mathrm{7}}}{\mathrm{392}}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{this}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense},\:\mathrm{it}'\mathrm{s}\:“\mathrm{cheaper}''\:\mathrm{to} \\ $$$$\mathrm{approximately}\:\mathrm{solve} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{check}\:\mathrm{the}\:\mathrm{solutions}\:\left(\mathrm{squaring}\:\mathrm{causes}\right. \\ $$$$\left.\mathrm{false}\:\mathrm{solutions}\right) \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{2}.\mathrm{73908}\:\bigstar \\ $$$${x}_{\mathrm{2}} \approx.\mathrm{200728}\:{wrong} \\ $$$${x}_{\mathrm{3}} \approx.\mathrm{399135}\:{wrong} \\ $$$${x}_{\mathrm{4}} \approx\mathrm{1}.\mathrm{13922}\:{wrong} \\ $$
Commented by mr W last updated on 03/Apr/21
thanks sirs!
$${thanks}\:{sirs}! \\ $$
Answered by mindispower last updated on 02/Apr/21
let Z=(√(x(x+1)))+(√(x(x+2)))  Y=x+(√((x+1)(x+2)))  Z^2 =2x^2 +3x+2x(√((x+1)(x+2)))  Y^2 =2x^2 +3x+2+2x(√((x+2)(x+1)))  Y^2 −Z^2 =2...1  Y+Z=2...2  (1)&(2)⇒Y−Z=1  ⇒Y=(3/2)  ⇔((3/2)−x)=(√((x+1)(x+2)))⇒−3x+(9/4)=3x+2  ⇒6x=(1/4)⇒x=(1/(24))....  (1/(24))... i will post complet solution later  i solved withe ⇒ not ⇔ must tchek if (1/(24)) is solution
$${let}\:{Z}=\sqrt{{x}\left({x}+\mathrm{1}\right)}+\sqrt{{x}\left({x}+\mathrm{2}\right)} \\ $$$${Y}={x}+\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$${Z}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}{x}\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$${Y}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}+\mathrm{2}{x}\sqrt{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)} \\ $$$${Y}^{\mathrm{2}} −{Z}^{\mathrm{2}} =\mathrm{2}…\mathrm{1} \\ $$$${Y}+{Z}=\mathrm{2}…\mathrm{2} \\ $$$$\left(\mathrm{1}\right)\&\left(\mathrm{2}\right)\Rightarrow{Y}−{Z}=\mathrm{1} \\ $$$$\Rightarrow{Y}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Leftrightarrow\left(\frac{\mathrm{3}}{\mathrm{2}}−{x}\right)=\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}\Rightarrow−\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{3}{x}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{6}{x}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{24}}…. \\ $$$$\frac{\mathrm{1}}{\mathrm{24}}…\:{i}\:{will}\:{post}\:{complet}\:{solution}\:{later} \\ $$$${i}\:{solved}\:{withe}\:\Rightarrow\:{not}\:\Leftrightarrow\:{must}\:{tchek}\:{if}\:\frac{\mathrm{1}}{\mathrm{24}}\:{is}\:{solution} \\ $$
Commented by mr W last updated on 02/Apr/21
thanks sir!  (1/(24)) is a soluton. but is it the only one?
$${thanks}\:{sir}! \\ $$$$\frac{\mathrm{1}}{\mathrm{24}}\:{is}\:{a}\:{soluton}.\:{but}\:{is}\:{it}\:{the}\:{only}\:{one}? \\ $$
Answered by liberty last updated on 03/Apr/21
for x≥0  (√x) ((√x) +(√(x+1)) )+(√(x+2)) ((√x) +(√(x+1)) )= 2  (√x) +(√(x+2)) = (2/( (√x) +(√(x+1))))  that can be written as  (√x) +(√(x+2)) = 2(√(x+1)) −2(√x)  3(√x) +(√(x+2)) = 2(√(x+1))  10x+2+6(√(x(x+2))) = 4x+4  3(√(x(x+2))) = 1−3x  9x^2 +18x = 9x^2 −6x+1  ⇔ 24x = 1 ; x=(1/(24))
$${for}\:{x}\geqslant\mathrm{0} \\ $$$$\sqrt{{x}}\:\left(\sqrt{{x}}\:+\sqrt{{x}+\mathrm{1}}\:\right)+\sqrt{{x}+\mathrm{2}}\:\left(\sqrt{{x}}\:+\sqrt{{x}+\mathrm{1}}\:\right)=\:\mathrm{2} \\ $$$$\sqrt{{x}}\:+\sqrt{{x}+\mathrm{2}}\:=\:\frac{\mathrm{2}}{\:\sqrt{{x}}\:+\sqrt{{x}+\mathrm{1}}} \\ $$$${that}\:{can}\:{be}\:{written}\:{as} \\ $$$$\sqrt{{x}}\:+\sqrt{{x}+\mathrm{2}}\:=\:\mathrm{2}\sqrt{{x}+\mathrm{1}}\:−\mathrm{2}\sqrt{{x}} \\ $$$$\mathrm{3}\sqrt{{x}}\:+\sqrt{{x}+\mathrm{2}}\:=\:\mathrm{2}\sqrt{{x}+\mathrm{1}} \\ $$$$\mathrm{10}{x}+\mathrm{2}+\mathrm{6}\sqrt{{x}\left({x}+\mathrm{2}\right)}\:=\:\mathrm{4}{x}+\mathrm{4} \\ $$$$\mathrm{3}\sqrt{{x}\left({x}+\mathrm{2}\right)}\:=\:\mathrm{1}−\mathrm{3}{x} \\ $$$$\mathrm{9}{x}^{\mathrm{2}} +\mathrm{18}{x}\:=\:\mathrm{9}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1} \\ $$$$\Leftrightarrow\:\mathrm{24}{x}\:=\:\mathrm{1}\:;\:{x}=\frac{\mathrm{1}}{\mathrm{24}} \\ $$

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