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Question Number 77885 by behi83417@gmail.com last updated on 11/Jan/20
solve for : x 1.(√((x−a)(x−b)))+(√((x−b)(x−c)))+(√((x−c)(x−a)))=d [a,b,c,d∈R try for: a=4,b=3,c=2,d=1] 2. (x−a^2 )(√(x−a))+(x−a)(√(x−a^2 ))=a^2 +a+1 3. (x−a^2 )(√(x^2 −a))+(x^2 −a)(√(x−a^2 ))=a^2 +a+1 [a∈R try for: a=(1/2) ]
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\::\:\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{1}.\sqrt{\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{b}}\right)}+\sqrt{\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{b}}\right)\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{c}}\right)}+\sqrt{\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)}=\boldsymbol{\mathrm{d}} \\ $$$$\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}},\boldsymbol{\mathrm{d}}\in\boldsymbol{\mathrm{R}}\right. \\ $$$$\left.\mathrm{try}\:\mathrm{for}:\:\:\mathrm{a}=\mathrm{4},\mathrm{b}=\mathrm{3},\mathrm{c}=\mathrm{2},\mathrm{d}=\mathrm{1}\right] \\ $$$$\mathrm{2}.\:\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}}+\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}+\mathrm{1} \\ $$$$\mathrm{3}.\:\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}+\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}\right)\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}+\mathrm{1} \\ $$$$\left[\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\right. \\ $$$$\left.\mathrm{try}\:\mathrm{for}:\:\mathrm{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\right] \\ $$$$ \\ $$
Answered by jagoll last updated on 12/Jan/20
1. (√((x−4)(x−3)))+(√((x−3)(x−2)))=1−(√((x−2)(x−4))) squaring (x−3){x−4+x−2}+2(x−3)(√((x−4)(x−2)))=1+(x−4)(x−2)−2(√((x−4)(x−2))) (x−3){2x−4+(√((x−4)(x−2))) }=1+(x−4)(x−2)−2(√((x−4)(x−2))) let u = x−4 ⇔ (u−1){2u+4+(√(u(u+2)))}=1+u(u+2)−2(√(u(u+2))) tobe continiu
$$\mathrm{1}.\:\sqrt{\left({x}−\mathrm{4}\right)\left({x}−\mathrm{3}\right)}+\sqrt{\left({x}−\mathrm{3}\right)\left({x}−\mathrm{2}\right)}=\mathrm{1}−\sqrt{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right)} \\ $$$${squaring} \\ $$$$\left({x}−\mathrm{3}\right)\left\{{x}−\mathrm{4}+{x}−\mathrm{2}\right\}+\mathrm{2}\left({x}−\mathrm{3}\right)\sqrt{\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)}=\mathrm{1}+\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)−\mathrm{2}\sqrt{\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)} \\ $$$$\left({x}−\mathrm{3}\right)\left\{\mathrm{2}{x}−\mathrm{4}+\sqrt{\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)}\:\right\}=\mathrm{1}+\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)−\mathrm{2}\sqrt{\left({x}−\mathrm{4}\right)\left({x}−\mathrm{2}\right)} \\ $$$${let}\:{u}\:=\:{x}−\mathrm{4}\: \\ $$$$\Leftrightarrow\:\left({u}−\mathrm{1}\right)\left\{\mathrm{2}{u}+\mathrm{4}+\sqrt{{u}\left({u}+\mathrm{2}\right)}\right\}=\mathrm{1}+{u}\left({u}+\mathrm{2}\right)−\mathrm{2}\sqrt{{u}\left({u}+\mathrm{2}\right)} \\ $$$${tobe}\:{continiu} \\ $$
Commented by mr W last updated on 12/Jan/20
dead−end street! it leads to an 8−th order equation which is not solvable.
$${dead}−{end}\:{street}! \\ $$$${it}\:{leads}\:{to}\:{an}\:\mathrm{8}−{th}\:{order}\:{equation}\:{which} \\ $$$${is}\:{not}\:{solvable}. \\ $$
Commented by jagoll last updated on 12/Jan/20
hahaha....thanks sir
$${hahaha}….{thanks}\:{sir} \\ $$
Commented by behi83417@gmail.com last updated on 12/Jan/20
dear master:mrW! thanks. I hope and wait for any try by you, sir.
$$\mathrm{dear}\:\mathrm{master}:\mathrm{mrW}!\:\mathrm{thanks}. \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{and}\:\mathrm{wait}\:\mathrm{for}\:\mathrm{any}\:\mathrm{try}\:\mathrm{by}\:\mathrm{you},\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 12/Jan/20
i don′t think a general solution (formula) is possible. assume a≥b≥c, we can say: if d<(√((a−c)×min(a−b,b−c))) ⇒no solution. if d≥(√((a−c)×min(a−b,b−c))) one or two solutions exist which lie in the range x≤c or x≥a. but a formula for the roots is not possible. as for the example a=4,b=3,c=2, (√((4−2)×min(1,1)))=(√2) therefore for d=1<(√2) there is no solution. see diagram. MJS sir has given the same statement.
$${i}\:{don}'{t}\:{think}\:{a}\:{general}\:{solution}\:\left({formula}\right) \\ $$$${is}\:{possible}. \\ $$$${assume}\:{a}\geqslant{b}\geqslant{c},\:{we}\:{can}\:{say}: \\ $$$${if}\:{d}<\sqrt{\left({a}−{c}\right)×{min}\left({a}−{b},{b}−{c}\right)}\:\Rightarrow{no}\:{solution}. \\ $$$${if}\:{d}\geqslant\sqrt{\left({a}−{c}\right)×{min}\left({a}−{b},{b}−{c}\right)} \\ $$$${one}\:{or}\:{two}\:{solutions}\:{exist}\:{which}\:{lie} \\ $$$${in}\:{the}\:{range}\:{x}\leqslant{c}\:{or}\:{x}\geqslant{a}.\:{but}\:{a}\:{formula} \\ $$$${for}\:{the}\:{roots}\:{is}\:{not}\:{possible}. \\ $$$${as}\:{for}\:{the}\:{example}\:{a}=\mathrm{4},{b}=\mathrm{3},{c}=\mathrm{2}, \\ $$$$\sqrt{\left(\mathrm{4}−\mathrm{2}\right)×{min}\left(\mathrm{1},\mathrm{1}\right)}=\sqrt{\mathrm{2}} \\ $$$${therefore}\:{for}\:{d}=\mathrm{1}<\sqrt{\mathrm{2}}\:{there}\:{is}\:{no}\:{solution}. \\ $$$${see}\:{diagram}. \\ $$$${MJS}\:{sir}\:{has}\:{given}\:{the}\:{same}\:{statement}. \\ $$
Commented by mr W last updated on 12/Jan/20
Commented by mr W last updated on 12/Jan/20
curve f(x) stands for f(x)=(√((x−a)(x−b)))+(√((x−b)(x−c)))+(√((x−c)(x−a))) and d_(min) =(√((a−c)×min(a−b,b−c))) solution for f(x)=d exists when d≥d_(min) .
$${curve}\:{f}\left({x}\right)\:{stands}\:{for} \\ $$$${f}\left({x}\right)=\sqrt{\left({x}−{a}\right)\left({x}−{b}\right)}+\sqrt{\left({x}−{b}\right)\left({x}−{c}\right)}+\sqrt{\left({x}−{c}\right)\left({x}−{a}\right)} \\ $$$${and} \\ $$$${d}_{{min}} =\sqrt{\left({a}−{c}\right)×{min}\left({a}−{b},{b}−{c}\right)} \\ $$$${solution}\:{for}\:{f}\left({x}\right)={d}\:{exists}\:{when} \\ $$$${d}\geqslant{d}_{{min}} . \\ $$
Commented by mr W last updated on 12/Jan/20
following is an example with a=5,b=3,c=2. for d=4: two solutions for d=2.5: one solution.
$${following}\:{is}\:{an}\:{example}\:{with} \\ $$$${a}=\mathrm{5},{b}=\mathrm{3},{c}=\mathrm{2}. \\ $$$${for}\:{d}=\mathrm{4}:\:{two}\:{solutions} \\ $$$${for}\:{d}=\mathrm{2}.\mathrm{5}:\:{one}\:{solution}. \\ $$
Commented by mr W last updated on 12/Jan/20
Commented by mr W last updated on 12/Jan/20
Commented by behi83417@gmail.com last updated on 12/Jan/20
thank you very much dear master. it is perfect sir. great solution by great man!
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{master}. \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{sir}. \\ $$$$\mathrm{great}\:\mathrm{solution}\:\mathrm{by}\:\mathrm{great}\:\mathrm{man}! \\ $$
Answered by MJS last updated on 12/Jan/20
1. let a≤b≤c; p≤q (x−p)(x−q)≥0 ⇒ x≤p∨x≥q ⇒ (x≤a∨x≥b)∧(x≤b∨x≥c)∧(x≤a∨x≥c) ⇒ x≤a∨x≥c ⇒ the minimum of (√((x−a)(x−b)))+(√((x−b)(x−c)))+(√((x−c)(x−a))) is either at x=a or x=c and thus its value is either (√((a−b)(a−c))) or (√((c−a)(c−b))) is (√((a−b)(a−c))) if a+c≥2b or (√((c−a)(c−b))) if a+c≤2b with a=4, b=3, c=2 the minimum is (√2) ⇒ no solution for d=1
$$\mathrm{1}. \\ $$$$\mathrm{let}\:{a}\leqslant{b}\leqslant{c};\:{p}\leqslant{q} \\ $$$$\left({x}−{p}\right)\left({x}−{q}\right)\geqslant\mathrm{0}\:\Rightarrow\:{x}\leqslant{p}\vee{x}\geqslant{q} \\ $$$$\Rightarrow \\ $$$$\left({x}\leqslant{a}\vee{x}\geqslant{b}\right)\wedge\left({x}\leqslant{b}\vee{x}\geqslant{c}\right)\wedge\left({x}\leqslant{a}\vee{x}\geqslant{c}\right) \\ $$$$\Rightarrow \\ $$$${x}\leqslant{a}\vee{x}\geqslant{c} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\sqrt{\left({x}−{a}\right)\left({x}−{b}\right)}+\sqrt{\left({x}−{b}\right)\left({x}−{c}\right)}+\sqrt{\left({x}−{c}\right)\left({x}−{a}\right)} \\ $$$$\mathrm{is}\:\mathrm{either}\:\mathrm{at}\:{x}={a}\:\mathrm{or}\:{x}={c}\:\mathrm{and}\:\mathrm{thus}\:\mathrm{its}\:\mathrm{value} \\ $$$$\mathrm{is}\:\mathrm{either}\:\sqrt{\left({a}−{b}\right)\left({a}−{c}\right)}\:\mathrm{or}\:\sqrt{\left({c}−{a}\right)\left({c}−{b}\right)} \\ $$$$\mathrm{is}\:\sqrt{\left({a}−{b}\right)\left({a}−{c}\right)}\:\mathrm{if}\:{a}+{c}\geqslant\mathrm{2}{b} \\ $$$$\mathrm{or}\:\sqrt{\left({c}−{a}\right)\left({c}−{b}\right)}\:\mathrm{if}\:{a}+{c}\leqslant\mathrm{2}{b} \\ $$$$ \\ $$$$\mathrm{with}\:{a}=\mathrm{4},\:{b}=\mathrm{3},\:{c}=\mathrm{2}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{d}=\mathrm{1} \\ $$
Commented by behi83417@gmail.com last updated on 12/Jan/20
dear proph :MJS! thank you very much sir.is there any chance to find: x in terms of :a,b,c,d?
$$\mathrm{dear}\:\mathrm{proph}\::\mathrm{MJS}!\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$$$\mathrm{sir}.\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{chance}\:\mathrm{to}\:\mathrm{find}:\:\mathrm{x}\:\mathrm{in} \\ $$$$\mathrm{terms}\:\mathrm{of}\::\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}? \\ $$

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