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 33569 by Joel578 last updated on 19/Apr/18

Given f(x) = x^3 + ax^2 + bx + c with a, b, c ∈ R, the roots are x_1 , x_2 , x_3 ∈ R Let λ is an positive integer that satisfied x_2 − x_1 = λ x_3 > (1/2)(x_1 + x_2 ) What is the max value of ((2a^3 + 27c − 9ab)/λ^3 ) ?

Givenf(x)=x3+ax2+bx+c witha,b,c∈R,therootsarex1,x2,x3∈R Letλisanpositiveintegerthatsatisfied x2−x1=λ x3>12(x1+x2) Whatisthemaxvalueof2a3+27c−9abλ3?

Answered by MJS last updated on 20/Apr/18

f(x)=(x−x_1 )(x−x_2 )(x−x_3 )= =x^3 −(x_1 +x_2 +x_3 )x^2 +(x_1 x_2 +x_1 x_3 +x_2 x_3 )x−x_1 x_2 x_3 a=−x_1 −x_2 −x_3 b=x_1 x_2 +x_1 x_3 +x_2 x_3 c=−x_1 x_2 x_3 x_2 =x_1 +λ a=−2x_1 −x_3 −λ b=x_1 ^2 +2x_1 x_3 +λ(x_1 +x_3 ) c=−x_1 ^2 x_3 −λx_1 x_3 ((2a^3 + 27c − 9ab)/λ^3 )= =−3((x_1 −x_3 )/λ)+3(((x_1 −x_3 )/λ))^2 +2(((x_1 −x_3 )/λ))^3 −2=t(λ) t′(λ)=3((x_1 −x_3 )/λ^2 )−6(((x_1 −x_3 )^2 )/λ^3 )−6(((x_1 −x_3 )^3 )/λ^4 ) t′(λ)=0 3(x_1 −x_3 )λ^2 −6(x_1 −x_3 )^2 λ−6(x_1 −x_3 )^3 =0 λ^2 −2(x_1 −x_3 )λ−2(x_1 −x_3 )^2 =0 λ=(x_1 −x_3 )±(√3)(x_1 −x_3 )=(x_1 −x_3 )(1±(√3)) x_3 > (1/2)(x_1 + x_2 ) ∧ x_2 =x_1 +λ ⇒ x_3 >x_1 +(λ/2) λ>0 ∧ x_3 >x_1 +(λ/2) ⇒ λ=(x_1 −x_3 )(1−(√3)) t(λ)=t((x_1 −x_3 )(1−(√3)))=((3(√3))/2) max(((2a^3 + 27c − 9ab)/λ^3 ))=((3(√3))/2)

f(x)=(x−x1)(x−x2)(x−x3)= =x3−(x1+x2+x3)x2+(x1x2+x1x3+x2x3)x−x1x2x3 a=−x1−x2−x3 b=x1x2+x1x3+x2x3 c=−x1x2x3 x2=x1+λ a=−2x1−x3−λ b=x12+2x1x3+λ(x1+x3) c=−x12x3−λx1x3 2a3+27c−9abλ3= =−3x1−x3λ+3(x1−x3λ)2+2(x1−x3λ)3−2=t(λ) t′(λ)=3x1−x3λ2−6(x1−x3)2λ3−6(x1−x3)3λ4 t′(λ)=0 3(x1−x3)λ2−6(x1−x3)2λ−6(x1−x3)3=0 λ2−2(x1−x3)λ−2(x1−x3)2=0 λ=(x1−x3)±3(x1−x3)=(x1−x3)(1±3) x3>12(x1+x2)∧x2=x1+λ⇒x3>x1+λ2 λ>0∧x3>x1+λ2⇒λ=(x1−x3)(1−3) t(λ)=t((x1−x3)(1−3))=332 max(2a3+27c−9abλ3)=332

Commented byJoel578 last updated on 20/Apr/18

thank you very much

thankyouverymuch

Terms of Service

Privacy Policy

Contact: info@tinkutara.com