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Question Number 205073 by mr W last updated on 07/Mar/24
if a, b, c are the roots of f(x)=x^3 −2024x^2 +2024x+2024 find (1/(1−a^2 ))+(1/(1−b^2 ))+(1/(1−c^2 ))=?
ifa,b,caretherootsoff(x)=x3−2024x2+2024x+2024find11−a2+11−b2+11−c2=?
Answered by cortano12 last updated on 07/Mar/24
= (a^2 /(1−a^2 )) +(b^2 /(1−b^2 )) +(c^2 /(1−c^2 )) + 3 = ((a^2 +b^2 −2(ab)^2 )/((1−a^2 )(1−b^2 ))) +(c^2 /(1−c^2 )) + 3 =((a^2 +b^2 +c^2 +3(abc)^2 −2((ab)^2 +(ac)^2 +(bc)^2 )/((1−a^2 )(1−b^2 )(1−c^2 )))+3
=a21−a2+b21−b2+c21−c2+3=a2+b2−2(ab)2(1−a2)(1−b2)+c21−c2+3=a2+b2+c2+3(abc)2−2((ab)2+(ac)2+(bc)2(1−a2)(1−b2)(1−c2)+3
Answered by Frix last updated on 07/Mar/24
y=(1/(1−x^2 )) ⇒ x=±(√((y−1)/y)) Inserting & transforming y^3 −((2027y^2 )/(2025))−((2021y)/(2025))−(1/(4100625))=0 ⇒ answer is ((2027)/(2025))
y=11−x2⇒x=±y−1yInserting&transformingy3−2027y22025−2021y2025−14100625=0⇒answeris20272025
Answered by A5T last updated on 07/Mar/24
WLOG ,let a=x+yi,b=x−yi,c=c where x,y,c∈R ?=(1/(1−(x+yi)^2 ))+(1/(1−(x−yi)^2 ))+(1/(1−c^2 )) ⇒2?=(1/(1−x−yi))+(1/(1−x+yi))+(1/(1−c))+(1/(1+x+yi))+(1/(1+x−yi))+(1/(1+c)) =((2(1−x))/((1−x)^2 +y^2 ))+(1/(1−c))+((2(1+x))/((1+x)^2 +y^2 ))+(1/(1+c)) =((2(1−c−x+xc)+1+x^2 −2x+y^2 )/(1+x^2 −2x+y^2 −c−cx^2 +2cx−cy^2 ))+((2(1+x+c+cx)+1+x^2 +2x+y^2 )/(1+x^2 +2x+y^2 +c+cx^2 +2cx+cy^2 )) =((3−2(c+2x)+x^2 +y^2 +2cx)/(1+x^2 +y^2 +2cx−(2x+c)−c(x^2 +y^2 )))+((3+x^2 +y^2 +2cx+2(2x+c))/(1+x^2 +y^2 +2cx+2x+c+c(x^2 +y^2 ))) =((3−(2024))/(2025))+((3×2025)/(2025))=((4054)/(2025))=2? ⇒?=((2027)/(2025)) [ab+bc+ca=2024⇒x^2 +y^2 +2cx=2024 abc=(x^2 +y^2 )c=−2024;a+b+c=2x+c=2024]
WLOG,leta=x+yi,b=x−yi,c=cwherex,y,c∈R?=11−(x+yi)2+11−(x−yi)2+11−c2⇒2?=11−x−yi+11−x+yi+11−c+11+x+yi+11+x−yi+11+c=2(1−x)(1−x)2+y2+11−c+2(1+x)(1+x)2+y2+11+c=2(1−c−x+xc)+1+x2−2x+y21+x2−2x+y2−c−cx2+2cx−cy2+2(1+x+c+cx)+1+x2+2x+y21+x2+2x+y2+c+cx2+2cx+cy2=3−2(c+2x)+x2+y2+2cx1+x2+y2+2cx−(2x+c)−c(x2+y2)+3+x2+y2+2cx+2(2x+c)1+x2+y2+2cx+2x+c+c(x2+y2)=3−(2024)2025+3×20252025=40542025=2?⇒?=20272025[ab+bc+ca=2024⇒x2+y2+2cx=2024abc=(x2+y2)c=−2024;a+b+c=2x+c=2024]
Answered by A5T last updated on 07/Mar/24
(1/(1−a))+(1/(1−b))+(1/(1−c))+(1/(1+a))+(1/(1+b))+(1/(1+c))=2? ((Σ(1−a)(1−b))/((1−a)(1−b)(1−c)))+((Σ(1+a)(1+b))/((1+a)(1+b)(1+c))) =((3−2(a+b+c)+ab+bc+ca)/(1−a−b−c+ab+ac+bc−abc))+((3+2(Σa)+Σab)/(1+Σa+Σab+abc)) =((3−2(2024)+2024)/(1−2024+2024+2024))+((3+2(2024)+2024)/(1+2024+2024−2024)) =((3−2024)/(2025))+((3×2025)/(2025))=((4054)/(2025))=2?⇒?=((2027)/(2025))
11−a+11−b+11−c+11+a+11+b+11+c=2?Σ(1−a)(1−b)(1−a)(1−b)(1−c)+Σ(1+a)(1+b)(1+a)(1+b)(1+c)=3−2(a+b+c)+ab+bc+ca1−a−b−c+ab+ac+bc−abc+3+2(Σa)+Σab1+Σa+Σab+abc=3−2(2024)+20241−2024+2024+2024+3+2(2024)+20241+2024+2024−2024=3−20242025+3×20252025=40542025=2?⇒?=20272025
Answered by mr W last updated on 07/Mar/24
(x−1+1)^3 −2024(x−1+1)^2 +2024(x−1+1)+2024=0 let t=x−1 (t+1)^3 −2024(t+1)^2 +2024(t+1)+2024=0 ((2025)/t^3 )−((2021)/t^2 )−((2021)/t)+1=0 ⇒Σ(1/(1−x))=−Σ(1/t)=−((2021)/(2025)) (x+1−1)^3 −2024(x+1−1)^2 +2024(x+1−1)+2024=0 let s=x+1 (s−1)^3 −2024(s−1)^2 +2024(s−1)+2024=0 s^3 −2027s^2 +6075s−2025=0 ((2025)/s^3 )−((6075)/s^2 )+((2027)/s)−1=0 ⇒Σ(1/(1+x))=Σ(1/s)=((6075)/(2025)) (1/(1−a^2 ))+(1/(1−b^2 ))+(1/(1−c^2 )) =(1/2)((1/(1−a))+(1/(1−b))+(1/(1−c))+(1/(1+a))+(1/(1+b))+(1/(1+c))) =(1/2)(Σ(1/(1−x))+Σ(1/(1+x))) =(1/2)(−((2021)/(2025))+((6075)/(2025)))=((2027)/(2025)) ✓
(x−1+1)3−2024(x−1+1)2+2024(x−1+1)+2024=0lett=x−1(t+1)3−2024(t+1)2+2024(t+1)+2024=02025t3−2021t2−2021t+1=0⇒Σ11−x=−Σ1t=−20212025(x+1−1)3−2024(x+1−1)2+2024(x+1−1)+2024=0lets=x+1(s−1)3−2024(s−1)2+2024(s−1)+2024=0s3−2027s2+6075s−2025=02025s3−6075s2+2027s−1=0⇒Σ11+x=Σ1s=6075202511−a2+11−b2+11−c2=12(11−a+11−b+11−c+11+a+11+b+11+c)=12(Σ11−x+Σ11+x)=12(−20212025+60752025)=20272025✓
Answered by pi314 last updated on 08/Mar/24
((P′(x))/(P(x)))=Σ_(a,b,c) (1/(x−c));((3x^2 −4048x+2024)/(x^3 −2024x^2 +2024x+2024)) S=(1/2)((1/(1−a))+(1/(1−b))+(1/(1−c)))+(1/2)((1/(1+a))+(1/(1+b))+(1/(1+c))) S=(1/2)(((P′(1))/(P(1)))−((P′(−1))/(P(−1))))=(1/2)(((−2021)/(2025))+((6075)/(2025)))=((2027)/(2025))
P′(x)P(x)=∑a,b,c1x−c;3x2−4048x+2024x3−2024x2+2024x+2024S=12(11−a+11−b+11−c)+12(11+a+11+b+11+c)S=12(P′(1)P(1)−P′(−1)P(−1))=12(−20212025+60752025)=20272025

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