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Roots of polynomials
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FM Roots of polynomials: Quartics
KUS objectives BAT Derive and use the relationships between the roots of quartic equations Starter: Solve π₯π₯π₯given that one root is π§=1βπ π₯π₯π₯ π₯=π₯π₯π₯=β2Β±π
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π π₯ 4 +π π₯ 3 +π π₯ 2 +ππ₯+π=0 has up to four roots .
Notes π π₯ 4 +π π₯ 3 +π π₯ 2 +ππ₯+π=0 has up to four roots . π π₯ 4 +π π₯ 3 +π π₯ 2 +π
π₯+π=π π₯βπΌ π₯βπ½ π₯βπΎ π₯βπΏ πΌ, π½, πΎ,πΏ are the roots =β¦ =π π₯ 4 βπ πΌ+π½+πΎ+πΏ π₯ 3 +π πΆπ·+πΆπΈ+πΆπΉ+π·πΈ+π·πΉ+πΈπΉ π₯ 2 π πΌπ½πΎ+πΌπ½πΏ+πΌπΎπΏ+π½πΎπΏ π₯+ππΌπ½πΎπΏ In shorthand πΌ= β π π And πΌπ½= π π And πΌπ½πΎ =β π π Equate the coefficients πΌ+π½+πΎ+πΏ =β π π And πΌπ½+πΌπΎ+πΌπΏ+π½πΎ+π½πΏ+πΎπΏ = π π And πΌπ½πΎ+πΌπ½πΏ+πΌπΎπΏ+π½πΎπΏ =β π π And πΌπ½πΎπΏ = π π
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WB C1a The quartic equation π₯ 4 +2 π₯ 3 +π π₯ 2 +ππ₯β60=0 has roots β,π½, πΎ, πΏ
Given that πΎ=β2+4π and Ξ΄= πΎ β a) show that β+π½β2= and that πΌπ½+3=0 Hence find all the roots of the quartic and the values of p and q πΌ+π½+πΎ+πΏ =β π π Gives πΌ+π½+ β2+4π + β2β4π =β2 Gives πΌ+π½β2=0 QED (1) πΌπ½πΎπΏ= π π Gives πΌπ½ β2+4π β2β4π =β (2) Substitute (1) into (2) Gives πΌ(2βπΌ) β2+4π β2β4π =β60 πΌ 2βπΌ (20)=β60 πΌ 2 β2πΌβ3=0 (πΌβ3)(πΌ+1)=0 so πΌ=3, β1 π½=β1, 3 So the roots are 3, β1, β2Β±4π
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Given that πΎ=β2+4π and Ξ΄= πΎ β a) show that β+π½β2=0 and that πΌπ½+3=0
WB C1b The quartic equation π₯ 4 +2 π₯ 3 +π π₯ 2 +ππ₯β60=0 has roots β,π½, πΎ, πΏ Given that πΎ=β2+4π and Ξ΄= πΎ β a) show that β+π½β2= and that πΌπ½+3=0 b) Hence find all the roots of the quartic and the values of p and q In shorthand πΌ= β π π And πΌπ½= π π And πΌπ½πΎ =β π π So the roots are 3, β1, β2+4π, β2β4π So πΌπ½+πΌπΎ+πΌπΏ+π½πΎ+π½πΏ+πΎπΏ = π π Gives π= β3 + β6+12π + β6β12π + 2β4π + 2+4π + 20 π=9 So πΌπ½πΎ+πΌπ½πΏ+πΌπΎπΏ+π½πΎπΏ =β π π gives βπ= 6β12π π β20 =52 So π=β52
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WB C The d NOW DO Ex 4C
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self-assess One thing learned is β One thing to improve is β
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