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4x-4-16x-3-24x-2-9x-1-0-using-any-method-find-real-value-of-x-that-satisfy-the-polynomial-

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Question Number 41606 by psyche-ace last updated on 10/Aug/18
4x^4 +16x^3 +24x^2 −9x−1=0 using any method. find real value of x that satisfy the polynomial
4\boldsymbolx4+16\boldsymbolx3+24\boldsymbolx2−9\boldsymbolx−1=0\boldsymbolusing\boldsymbolany\boldsymbolmethod.\boldsymbolfind\boldsymbolreal\boldsymbolvalue\boldsymbolof\boldsymbolx\boldsymbolthat\boldsymbolsatisfy\boldsymbolthe\boldsymbolpolynomial
Answered by MJS last updated on 10/Aug/18
f(x)=x^4 +4x^3 +6x^2 −(9/4)x−(1/4)=0 f′(x)=4x^3 +12x^2 +12x−(9/4) x^3 +3x^2 +3x−(9/(16))=0 x=t−1 t^3 −((25)/(16))=0 ⇒ t=(((100))^(1/3) /4) ⇒ x=((−4+((100))^(1/3) )/4) f′(x) has got only 1 real root ⇒ f(x) has 0 or 2 real roots f′′(x)=12x^2 +24x+12 f(((−4+((100))^(1/3) )/4))=(5/(64))(64−15((100))^(1/3) )≈−.44 f′′(((−4+((100))^(1/3) )/4))=((15((10))^(1/3) )/2)>0 ⇒ f(x) has 2 real roots let f(x)=(x−a−(√b))(x−a+(√b))(x−c−(√d)i)(x−c+(√d)i) ⇒ a+c+2=0 a^2 +4ac−b+c^2 +d−6=0 8a^2 c+8ac^2 +8ad−8bc−9=0 a^2 c^2 +a^2 d−bc^2 −bd+(1/4)=0 a=−c−2 −b−2c^2 −4c+d−2=0 −8bc+16c^2 −8cd+32c−16d−9=0 −bc^2 −bd+c^4 +4c^3 +c^2 d+4c^2 +4cd+4d+(1/4)=0 a=−c−2 b=−2c^2 −4c+d−2 16c^3 +48c^2 −16cd+48c−16d−9=0 3c^4 +8c^3 +2c^2 d+6c^2 +8cd−d^2 +6d+(1/4)=0 a=−c−2 b=−2c^2 −4c+d−2 d=((16c^3 +48c^2 +48c−9)/(16(c+1))) 4c^4 +16c^3 +24c^2 +16c−1−((625)/(256(c+1)^2 ))=0 c^6 +6c^5 +15c^4 +20c^3 +((55)/4)c^2 +(7/2)c−((881)/(1024))=0 c=t−1 t^6 −(5/4)t^2 −((625)/(1024))=0 t=(√z) z^3 −(5/4)z−((625)/(1024))=0 z=(1/(48))(((30(1125+(√(282585))))^(1/3) +((30(1125−(√(282585))))^(1/3) )≈ ≈1.30994 t≈1.14453 c≈.144525 a=−2.14453 b=−2.67513 d=−.055256 a±(√b)=−2.14453±1.63558i c+(√d)i=−.0905405 c−(√d)i=.379591
f(x)=x4+4x3+6x2−94x−14=0f′(x)=4x3+12x2+12x−94x3+3x2+3x−916=0x=t−1t3−2516=0⇒t=10034⇒x=−4+10034f′(x)hasgotonly1realroot⇒f(x)has0or2realrootsf″(x)=12x2+24x+12f(−4+10034)=564(64−151003)≈−.44f″(−4+10034)=151032>0⇒f(x)has2realrootsletf(x)=(x−a−b)(x−a+b)(x−c−di)(x−c+di)⇒a+c+2=0a2+4ac−b+c2+d−6=08a2c+8ac2+8ad−8bc−9=0a2c2+a2d−bc2−bd+14=0a=−c−2−b−2c2−4c+d−2=0−8bc+16c2−8cd+32c−16d−9=0−bc2−bd+c4+4c3+c2d+4c2+4cd+4d+14=0a=−c−2b=−2c2−4c+d−216c3+48c2−16cd+48c−16d−9=03c4+8c3+2c2d+6c2+8cd−d2+6d+14=0a=−c−2b=−2c2−4c+d−2d=16c3+48c2+48c−916(c+1)4c4+16c3+24c2+16c−1−625256(c+1)2=0c6+6c5+15c4+20c3+554c2+72c−8811024=0c=t−1t6−54t2−6251024=0t=zz3−54z−6251024=0z=148(30(1125+2825853+30(1125−2825853)≈≈1.30994t≈1.14453c≈.144525a=−2.14453b=−2.67513d=−.055256a±b=−2.14453±1.63558ic+di=−.0905405c−di=.379591
Commented by MJS last updated on 10/Aug/18
to eliminate the bx^(n−1) you can always set x=t−(b/n) if the constant factor if ax^n is equal to 1 (a=1) examples: 3x^2 +4x−5=0 x^2 +(4/3)x−(5/3)=0 x=t−(2/3) (t−(2/3))^2 +(4/3)(t−(2/3))−(5/3)=0 t^2 −((19)/9)=0 (which in this case is the same as the usual formula x=−(p/2)±(√((p^2 /4)−q)) 4x^3 −5x^2 +3x−2=0 x^3 −(5/4)x^2 +(3/4)x−(1/2)=0 x=t+(5/(12)) (t+(5/(12)))^3 −(5/4)(t+(5/(12)))^2 +(3/4)(t+(5/(12)))−(1/2)=0 t^3 +((11)/(48))t−((287)/(864))=0 (which in this case leads to Cardano′s formula or in case of 3 real solutions to the trigonometric method) in case of a 4^(th) −degree polynome with 2 real and 2 complex roots my method leads to a polynome of 6^(th) degree which can be reduced to one of 3^(rd) degree... will post an example later
toeliminatethebxn−1youcanalwayssetx=t−bniftheconstantfactorifaxnisequalto1(a=1)examples:3x2+4x−5=0x2+43x−53=0x=t−23(t−23)2+43(t−23)−53=0t2−199=0(whichinthiscaseisthesameastheusualformulax=−p2±p24−q4x3−5x2+3x−2=0x3−54x2+34x−12=0x=t+512(t+512)3−54(t+512)2+34(t+512)−12=0t3+1148t−287864=0(whichinthiscaseleadstoCardano′sformulaorincaseof3realsolutionstothetrigonometricmethod)incaseofa4th−degreepolynomewith2realand2complexrootsmymethodleadstoapolynomeof6thdegreewhichcanbereducedtooneof3rddegree…willpostanexamplelater
Commented by Tawa1 last updated on 10/Aug/18
Sir how can we know that x = t − 1
Sirhowcanweknowthatx=t−1
Commented by Tawa1 last updated on 10/Aug/18
wow, please send it sir. God bless you sir
wow,pleasesenditsir.Godblessyousir
Commented by Tawa1 last updated on 10/Aug/18
Am still expecting the other method and examples sir. God bless you
Amstillexpectingtheothermethodandexamplessir.Godblessyou
Commented by Tawa1 last updated on 10/Aug/18
i will love to see the formulars sir
iwilllovetoseetheformularssir
Commented by Tawa1 last updated on 11/Aug/18
Sir MJS am still expecting sir
SirMJSamstillexpectingsir
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
4x^2 (x^2 +4x+4)+8x^2 −9x−1=0 {(2x)(x+2)}^2 +9x^2 −6x+1−x^2 −3x−2=0 {(2x)(x+2)}^2 +(3x−1)^2 −(x^2 +3x+2)=0 {(2x)(x+2)}^2 +(3x−1)^2 −{x^2 +2.x.(3/2)+(9/4)+2−(9/4)}=0 {(2x)(x+2)}^2 +(3x−1)^2 −{(x+(3/2))^2 −(1/4)}=0 [{(2x)(x+2)}^2 +(3x−1)^2 +(1/4)]−[(x+(3/2))^2 ]=0 now i want to say some thing the value of [{(2x)(x+2)}^2 +(3x−1)^2 +(1/4)] is +ve and say its value is p now the value of [(x+(3/2))^2 ] is also positve and say value is q now the eqn is p−q=0 that imply the value of p should be equal to q to make the equation true but {(2x)(x+2)}^2 >[(x+(3/2))^2 ] so to me p can not be equal to q to make the equation true.. to me p>>q so i think this equation is not feasible...
4x2(x2+4x+4)+8x2−9x−1=0{(2x)(x+2)}2+9x2−6x+1−x2−3x−2=0{(2x)(x+2)}2+(3x−1)2−(x2+3x+2)=0{(2x)(x+2)}2+(3x−1)2−{x2+2.x.32+94+2−94}=0{(2x)(x+2)}2+(3x−1)2−{(x+32)2−14}=0[{(2x)(x+2)}2+(3x−1)2+14]−[(x+32)2]=0nowiwanttosaysomethingthevalueof[{(2x)(x+2)}2+(3x−1)2+14]is+veandsayitsvalueispnowthevalueof[(x+32)2]isalsopositveandsayvalueisqnowtheeqnisp−q=0thatimplythevalueofpshouldbeequaltoqtomaketheequationtruebut{(2x)(x+2)}2>[(x+32)2]sotomepcannotbeequaltoqtomaketheequationtrue..tomep>>qsoithinkthisequationisnotfeasible…
Commented by Tawa1 last updated on 10/Aug/18
God bless you sir
GodblessyousirGodblessyousir

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