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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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