Menu Close

x-3-ax-2-bx-c-0-Transform-to-t-3-k-0-

Tinku Tara 0 Comments
Pin




Question Number 56467 by ajfour last updated on 17/Mar/19
x^3 +ax^2 +bx+c=0 Transform to t^3 +k=0 .
x3+ax2+bx+c=0Transformtot3+k=0.
Answered by ajfour last updated on 17/Mar/19
let x=((pt+q)/(t+1)) ⇒ (pt+q)^3 +a(pt+q)^2 (t+1) +b(pt+q)(t+1)^2 +c(t+1)^3 =0 (p^3 +ap^2 +bp+c)t^3 +(3p^2 q+ap^2 +2apq+bq+2bp+3c)t^2 +(3pq^2 +2apq+aq^2 +bp+2bq+3c)t +q^3 +aq^2 +bq+c=0 Let 3p^2 q+ap^2 +2apq+bq+2bp+3c=0 & 3pq^2 +2apq+aq^2 +bp+2bq+3c=0 subtracting 3pq+a(p+q)−b+2b=0 ⇒ 3pq+a(p+q)+b=0 (i) using this −p[a(p+q)+b]+ap^2 +2apq+bq +2bp+3c = 0 ⇒ apq+bq+bp+3c=0 ⇒ apq+b(p+q)+3c=0 (ii) ⇒ p+q=( determinant ((3,(−b)),(a,(−3c)))/ determinant ((3,a),(a,b)))=((−9c+ab)/(3b−a^2 )) =s ⇒ pq=( determinant (((−b),a),((−3c),b))/ determinant ((3,a),(a,b)))=((−b^2 +3ac)/(3b−a^2 )) =m ⇒ p,q are roots of eq. z^2 −sz+m = 0 z=(s/2)±(√((s^2 /4)−m)) So (p^3 +ap^2 +bp+c)t^3 +q^3 +aq^2 +bq+c = 0 t=−(((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bp+c)))^(1/3) x=((pt+q)/(t+1)) .
letx=pt+qt+1⇒(pt+q)3+a(pt+q)2(t+1)+b(pt+q)(t+1)2+c(t+1)3=0(p3+ap2+bp+c)t3+(3p2q+ap2+2apq+bq+2bp+3c)t2+(3pq2+2apq+aq2+bp+2bq+3c)t+q3+aq2+bq+c=0Let3p2q+ap2+2apq+bq+2bp+3c=0&3pq2+2apq+aq2+bp+2bq+3c=0subtracting3pq+a(p+q)−b+2b=0⇒3pq+a(p+q)+b=0(i)usingthis−p[a(p+q)+b]+ap2+2apq+bq+2bp+3c=0⇒apq+bq+bp+3c=0⇒apq+b(p+q)+3c=0(ii)⇒p+q=|3−ba−3c||3aab|=−9c+ab3b−a2=s⇒pq=|−ba−3cb||3aab|=−b2+3ac3b−a2=m⇒p,qarerootsofeq.z2−sz+m=0z=s2±s24−mSo(p3+ap2+bp+c)t3+q3+aq2+bq+c=0t=−(q3+aq2+bq+cp3+ap2+bp+c)1/3x=pt+qt+1.
Commented by behi83417@gmail.com last updated on 17/Mar/19
let try for: [x^3 +2x^2 +3x+4=0] {a=2,b=3,c=4} p+q=((−9c+ab)/(3b−a^2 ))=((−9×4+2×3)/(9−4))=−6 pq=((−b^2 +3ac)/(3b−a^2 ))=((−9+3×2×4)/5)=+3 ⇒z^2 +6z+3=0⇒z=((−6±(√(36−12)))/2)= ⇒z=−3±(√6)⇒ { ((p=−3+(√6)=−0.55,q=−3−(√6)=−5.44)),((p=−3−(√6),q=−3+(√6))) :} t^3 =−((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bp+c))=−(((−5.44)^3 +2(−5.44)^2 +3(−5.44)+4)/((−0.55)^3 +2(−0.55)^2 +3(−0.55)+4))=40.924 t=(40.925)^(1/3) =3.446 x=((pt+q)/(t+1))=((−0.55×3.446−5.44)/(3.446+1))=−1.65[ok!]
lettryfor:[x3+2x2+3x+4=0]{a=2,b=3,c=4}p+q=−9c+ab3b−a2=−9×4+2×39−4=−6pq=−b2+3ac3b−a2=−9+3×2×45=+3⇒z2+6z+3=0⇒z=−6±36−122=⇒z=−3±6⇒{p=−3+6=−0.55,q=−3−6=−5.44p=−3−6,q=−3+6t3=−q3+aq2+bq+cp3+ap2+bp+c=−(−5.44)3+2(−5.44)2+3(−5.44)+4(−0.55)3+2(−0.55)2+3(−0.55)+4=40.924t=(40.925)13=3.446x=pt+qt+1=−0.55×3.446−5.443.446+1=−1.65[ok!]
Commented by mr W last updated on 17/Mar/19
��������
Commented by behi83417@gmail.com last updated on 17/Mar/19
Commented by behi83417@gmail.com last updated on 17/Mar/19
congratulations sir Ajfour! this is a perfect way for 3rd deg.eq thanks for sharing this knowledges.
congratulationssirAjfour!thisisaperfectwayfor3rddeg.eqthanksforsharingthisknowledges.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *