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Question Number 202250 by MATHEMATICSAM last updated on 23/Dec/23

(a^2 − bc)x^2 + 2(b^2 − ca)x + (c^2 − ab) = 0 has two equal roots. Show that either b = 0 or (a^2 /(bc)) + (b^2 /(ca)) + (c^2 /(ab)) = 3.

(a2−bc)x2+2(b2−ca)x+(c2−ab)=0hastwoequalroots.Showthateitherb=0ora2bc+b2ca+c2ab=3.

Answered by esmaeil last updated on 23/Dec/23

r_1 −r_2 =^(r_1 =r_2 ) ((√δ^( ′) )/(∣a∣))=0→ ((√((b^2 −ca)^2 −(c^2 −ab)(a^2 −bc)))/(a^2 −bc))=0→ b^4 +c^2 a^2 −2b^2 ac−c^2 a^2 +bc^3 +ba^3 −ab^2 c =0 →b^4 −3b^2 ac+bc^3 +ba^3 ==→ b(b^3 −3bac+c^3 +a^3 )=0→ b=0 or (b^3 /(abc))−((3abc)/(abc))+(c^3 /(abc))+((ba^3 )/(abc))=0→ (b^2 /(ac))+(c^2 /(ab))+(a^2 /(bc))=3

r1−r2=r1=r2δ′∣a∣=0→(b2−ca)2−(c2−ab)(a2−bc)a2−bc=0→b4+c2a2−2b2ac−c2a2+bc3+ba3−ab2c=0→b4−3b2ac+bc3+ba3==→b(b3−3bac+c3+a3)=0→b=0orb3abc−3abcabc+c3abc+ba3abc=0→b2ac+c2ab+a2bc=3

Answered by Rasheed.Sindhi last updated on 23/Dec/23

α & α are the roots α+α=2α=−((2(b^2 − ca))/((a^2 − bc)))⇒α=((b^2 − ca)/(a^2 − bc)) α^2 =(((b^2 − ca)/(a^2 − bc)))^2 =((c^2 − ab)/(a^2 − bc)) (((b^2 − ca)^2 )/(a^2 − bc))=c^2 − ab (b^2 − ca)^2 =(a^2 − bc)(c^2 − ab) b^4 −2ab^2 c+c^2 a^2 =c^2 a^2 −a^3 b−bc^3 +ab^2 c b^4 −3ab^2 c=−a^3 b−bc^3 b^4 −3ab^2 c+a^3 b+bc^3 =0 b(b^3 −3abc+a^3 +c^3 )=0 b=0✓ ∣ a^3 +b^3 +c^3 =3abc ((a^3 +b^3 +c^3 )/(abc))=3 (a^2 /(bc)) + (b^2 /(ca)) + (c^2 /(ab)) = 3 ✓

α&αaretherootsα+α=2α=−2(b2−ca)(a2−bc)⇒α=b2−caa2−bcα2=(b2−caa2−bc)2=c2−aba2−bc(b2−ca)2a2−bc=c2−ab(b2−ca)2=(a2−bc)(c2−ab)b4−2ab2c+c2a2=c2a2−a3b−bc3+ab2cb4−3ab2c=−a3b−bc3b4−3ab2c+a3b+bc3=0b(b3−3abc+a3+c3)=0b=0✓∣a3+b3+c3=3abca3+b3+c3abc=3a2bc+b2ca+c2ab=3✓

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