Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 203474 by ajfour last updated on 20/Jan/24

((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((z+1)/(z−1)) Find z∈R.

z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=z+1z−1Findz∈R.

Answered by Rasheed.Sindhi last updated on 20/Jan/24

Using Componendo-Dividendo (((z^4 +4z^3 −6z^2 −4z+1)+(z^4 −4z^3 −6z^2 +4z+1))/((z^4 +4z^3 −6z^2 −4z+1)−(z^4 −4z^3 −6z^2 +4z+1))) =(((z+1)+(z−1))/((z+1)−(z−1))) ((2z^4 −12z^2 +2)/(8z^3 −8z))=((2z)/2) ((z^4 −6z^2 +1)/(4z^3 −4z))=z z^4 −6z^2 +1=4z^4 −4z^2 3z^4 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 z^2 =−1,(1/3) z=±i, ±(1/( (√3))) ✓

UsingComponendo−Dividendo(z4+4z3−6z2−4z+1)+(z4−4z3−6z2+4z+1)(z4+4z3−6z2−4z+1)−(z4−4z3−6z2+4z+1)=(z+1)+(z−1)(z+1)−(z−1)2z4−12z2+28z3−8z=2z2z4−6z2+14z3−4z=zz4−6z2+1=4z4−4z23z4+2z2−1=0(z2+1)(3z2−1)=0z2=−1,13z=±i,±13✓

Answered by Rasheed.Sindhi last updated on 20/Jan/24

Using Cross Multiplication ▶z^5 +4z^4 −6z^3 −4z^2 +z−z^4 −4z^3 +6z^2 +4z−1 =z^5 −4z^4 −6z^3 +4z^2 +z+z^4 −4z^3 −6z^2 +4z+1 ▶8z^4 −2z^4 +4z^2 −2=0 ▶3z^4 +2z^2 −1=0 ▶(z^2 +1)(3z^2 −1)=0 ▶ z^2 =−1,(1/3) ▶z=±i, ±(1/( (√3))) ✓

UsingCrossMultiplication▸z5+4z4−6z3−4z2+z−z4−4z3+6z2+4z−1=z5−4z4−6z3+4z2+z+z4−4z3−6z2+4z+1▸8z4−2z4+4z2−2=0▸3z4+2z2−1=0▸(z2+1)(3z2−1)=0▸z2=−1,13▸z=±i,±13✓

Answered by Rasheed.Sindhi last updated on 20/Jan/24

((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((k(z+1))/(k(z−1))) (let) where k is such that: { ((z^4 +4z^3 −6z^2 −4z+1=k(z+1))),(( &)),((z^4 −4z^3 −6z^2 +4z+1=k(z−1))) :} { ((z^4 +4z^3 −6z^2 −4z+1−kz−k=0)),((z^4 −4z^3 −6z^2 +4z+1−kz+k=0)) :} Adding & Subtracting: { ((z^4 −6z^2 −kz+1=0)),((4z^3 −4z−k=0⇒k=4z^3 −4z)) :} ⇒z^4 −6z^2 −(4z^3 −4z)z+1=0 z^4 −6z^2 −4z^4 +4z^2 +1=0 3z^4 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 z^2 =−1,(1/3) z=±i, ±((√3)/3) ✓

z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=k(z+1)k(z−1)(let)wherekissuchthat:{z4+4z3−6z2−4z+1=k(z+1)&z4−4z3−6z2+4z+1=k(z−1){z4+4z3−6z2−4z+1−kz−k=0z4−4z3−6z2+4z+1−kz+k=0Adding&Subtracting:{z4−6z2−kz+1=04z3−4z−k=0⇒k=4z3−4z⇒z4−6z2−(4z3−4z)z+1=0z4−6z2−4z4+4z2+1=03z4+2z2−1=0(z2+1)(3z2−1)=0z2=−1,13z=±i,±33✓

Answered by Rasheed.Sindhi last updated on 20/Jan/24

Using k-method ((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((z+1)/(z−1))=k (let) z+1=kz−k kz−z=k+1 z=((k+1)/(k−1)) ((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=k z^4 +4z^3 −6z^2 −4z+1−k(z^4 −4z^3 −6z^2 +4z+1)=0 z^4 +4z^3 −6z^2 −4z+1−kz^4 +4kz^3 +6kz^2 −4kz−k=0 (1−k)z^4 +4(1+k)z^3 −6(1−k)z^2 −4(1+k)z+1−k=0 z^4 +4(((1+k)/(1−k)))z^3 −6z^2 −4(((1+k)/(1−k)))z+1=0 z^4 −4(((k+1)/(k−1)))z^3 −6z^2 +4(((k+1)/(k−1)))z+1=0 z^4 −4(z)z^3 −6z^2 +4(z)z+1=0 −3z^3 −2z^2 +1=0 3z^3 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 z^2 =−1,(1/3) z=±i, ±((√3)/3)

Usingk−methodz4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=z+1z−1=k(let)z+1=kz−kkz−z=k+1z=k+1k−1z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=kz4+4z3−6z2−4z+1−k(z4−4z3−6z2+4z+1)=0z4+4z3−6z2−4z+1−kz4+4kz3+6kz2−4kz−k=0(1−k)z4+4(1+k)z3−6(1−k)z2−4(1+k)z+1−k=0z4+4(1+k1−k)z3−6z2−4(1+k1−k)z+1=0z4−4(k+1k−1)z3−6z2+4(k+1k−1)z+1=0z4−4(z)z3−6z2+4(z)z+1=0−3z3−2z2+1=03z3+2z2−1=0(z2+1)(3z2−1)=0z2=−1,13z=±i,±33

Answered by Rasheed.Sindhi last updated on 29/Jan/24

Componedo-Dividendo version 2 ((z^4 +4z^3 −6z^2 −4z+1)/(z^4 −4z^3 −6z^2 +4z+1))=((z+1)/(z−1)) (((z^4 −6z^2 +1)+(4z^3 −4z))/((z^4 −6z^2 +1)−(4z^3 −4z)))=(((z)+(1))/((z)−(1))) determinant (((((a+b)/(a−b))=((c+d)/(c−d))⇔(a/b)=(c/d)))) ((z^4 −6z^2 +1)/(4z^3 −4z))=(z/1) z^4 −6z^2 +1=4z^4 −4z^2 3z^4 +2z^2 −1=0 (z^2 +1)(3z^2 −1)=0 { ((z^2 +1=0 ⇒z∉R)),((3z^2 −1=0⇒z=±(1/( (√3))))) :}

Componedo−Dividendoversion2z4+4z3−6z2−4z+1z4−4z3−6z2+4z+1=z+1z−1(z4−6z2+1)+(4z3−4z)(z4−6z2+1)−(4z3−4z)=(z)+(1)(z)−(1)a+ba−b=c+dc−d⇔ab=cdz4−6z2+14z3−4z=z1z4−6z2+1=4z4−4z23z4+2z2−1=0(z2+1)(3z2−1)=0{z2+1=0⇒z∉R3z2−1=0⇒z=±13

Commented by ajfour last updated on 20/Jan/24

Thanks sir. Good ways.

Thankssir.Goodways.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com