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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.
$$\left({a}^{\mathrm{2}} \:−\:{bc}\right){x}^{\mathrm{2}} \:+\:\mathrm{2}\left({b}^{\mathrm{2}} \:−\:{ca}\right){x}\:+\:\left({c}^{\mathrm{2}} \:−\:{ab}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{roots}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{either}\: \\ $$$${b}\:=\:\mathrm{0}\:\mathrm{or}\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:\mathrm{3}. \\ $$
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
$$\underset{\mathrm{1}} {{r}}−{r}_{\mathrm{2}} \overset{{r}_{\mathrm{1}} =\underset{\mathrm{2}} {{r}}} {=}\frac{\sqrt{\delta^{\:'} }}{\mid{a}\mid}=\mathrm{0}\rightarrow \\ $$$$\frac{\sqrt{\left({b}^{\mathrm{2}} −{ca}\right)^{\mathrm{2}} −\left({c}^{\mathrm{2}} −{ab}\right)\left({a}^{\mathrm{2}} −{bc}\right)}}{{a}^{\mathrm{2}} −{bc}}=\mathrm{0}\rightarrow \\ $$$${b}^{\mathrm{4}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {ac}−{c}^{\mathrm{2}} {a}^{\mathrm{2}} +{bc}^{\mathrm{3}} +{ba}^{\mathrm{3}} −{ab}^{\mathrm{2}} {c} \\ $$$$=\mathrm{0} \\ $$$$\rightarrow{b}^{\mathrm{4}} −\mathrm{3}{b}^{\mathrm{2}} {ac}+{bc}^{\mathrm{3}} +{ba}^{\mathrm{3}} ==\rightarrow \\ $$$${b}\left({b}^{\mathrm{3}} −\mathrm{3}{bac}+{c}^{\mathrm{3}} +{a}^{\mathrm{3}} \right)=\mathrm{0}\rightarrow \\ $$$${b}=\mathrm{0}\:{or}\:\frac{{b}^{\mathrm{3}} }{{abc}}−\frac{\mathrm{3}{abc}}{{abc}}+\frac{{c}^{\mathrm{3}} }{{abc}}+\frac{{ba}^{\mathrm{3}} }{{abc}}=\mathrm{0}\rightarrow \\ $$$$\frac{{b}^{\mathrm{2}} }{{ac}}+\frac{{c}^{\mathrm{2}} }{{ab}}+\frac{{a}^{\mathrm{2}} }{{bc}}=\mathrm{3} \\ $$
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 ✓
$$\alpha\:\&\:\alpha\:{are}\:{the}\:{roots} \\ $$$$\alpha+\alpha=\mathrm{2}\alpha=−\frac{\mathrm{2}\left({b}^{\mathrm{2}} \:−\:{ca}\right)}{\left({a}^{\mathrm{2}} \:−\:{bc}\right)}\Rightarrow\alpha=\frac{{b}^{\mathrm{2}} \:−\:{ca}}{{a}^{\mathrm{2}} \:−\:{bc}} \\ $$$$\alpha^{\mathrm{2}} =\left(\frac{{b}^{\mathrm{2}} \:−\:{ca}}{{a}^{\mathrm{2}} \:−\:{bc}}\right)^{\mathrm{2}} =\frac{{c}^{\mathrm{2}} \:−\:{ab}}{{a}^{\mathrm{2}} \:−\:{bc}} \\ $$$$\:\:\:\:\:\:\frac{\left({b}^{\mathrm{2}} \:−\:{ca}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} \:−\:{bc}}={c}^{\mathrm{2}} \:−\:{ab} \\ $$$$\:\:\:\:\:\left({b}^{\mathrm{2}} \:−\:{ca}\right)^{\mathrm{2}} =\left({a}^{\mathrm{2}} \:−\:{bc}\right)\left({c}^{\mathrm{2}} \:−\:{ab}\right) \\ $$$$\:{b}^{\mathrm{4}} −\mathrm{2}{ab}^{\mathrm{2}} {c}+{c}^{\mathrm{2}} {a}^{\mathrm{2}} ={c}^{\mathrm{2}} {a}^{\mathrm{2}} −{a}^{\mathrm{3}} {b}−{bc}^{\mathrm{3}} +{ab}^{\mathrm{2}} {c} \\ $$$$\:\:{b}^{\mathrm{4}} −\mathrm{3}{ab}^{\mathrm{2}} {c}=−{a}^{\mathrm{3}} {b}−{bc}^{\mathrm{3}} \\ $$$$\:\:{b}^{\mathrm{4}} −\mathrm{3}{ab}^{\mathrm{2}} {c}+{a}^{\mathrm{3}} {b}+{bc}^{\mathrm{3}} =\mathrm{0} \\ $$$${b}\left({b}^{\mathrm{3}} −\mathrm{3}{abc}+{a}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$${b}=\mathrm{0}\checkmark\:\mid\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:\mathrm{3}\:\checkmark \\ $$

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