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If-two-roots-of-ax-2-bx-c-0-are-and-then-1-a-2-c-2-1-a-2-c-2-

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Question Number 205502 by MATHEMATICSAM last updated on 22/Mar/24
If two roots of ax^2 + bx + c = 0 are α and β then (1/((aα^2 + c)^2 )) + (1/((aβ^2 + c)^2 )) = ?
Iftworootsofax2+bx+c=0areαandβthen1(aα2+c)2+1(aβ2+c)2=?
Answered by Rasheed.Sindhi last updated on 23/Mar/24
Another Way (1/((aα^2 + c)^2 )) + (1/((aβ^2 + c)^2 )) =(1/(a^2 (α^2 + (c/a))^2 )) + (1/(a^2 (β^2 + (c/a))^2 )) =(1/(a^2 (α^2 + αβ)^2 )) + (1/(a^2 (β^2 + αβ)^2 )) =(1/(a^2 α^2 (α + β)^2 )) + (1/(a^2 β^2 (β + α)^2 )) =(1/(a^2 (α+β)^2 ))((1/α^2 )+(1/β^2 )) =(1/(a^2 (α+β)^2 ))∙((α^2 +β^2 )/((αβ)^2 )) =(1/(a^2 (α+β)^2 ))∙(((α+β)^2 −2αβ)/((αβ)^2 )) =(1/(a^2 (−(b/a))^2 ))∙(((−(b/a))^2 −2((c/a)))/(((c/a))^2 )) =(1/b^2 )∙(((b^2 /a^2 )−((2c)/a))/(c^2 /a^2 )) =(1/b^2 )∙(((b^2 −2ac)/a^2 )/(c^2 /a^2 )) =((b^2 −2ac)/(b^2 c^2 ))
AnotherWay1(aα2+c)2+1(aβ2+c)2=1a2(α2+ca)2+1a2(β2+ca)2=1a2(α2+αβ)2+1a2(β2+αβ)2=1a2α2(α+β)2+1a2β2(β+α)2=1a2(α+β)2(1α2+1β2)=1a2(α+β)2⋅α2+β2(αβ)2=1a2(α+β)2⋅(α+β)2−2αβ(αβ)2=1a2(−ba)2⋅(−ba)2−2(ca)(ca)2=1b2⋅b2a2−2cac2a2=1b2⋅b2−2aca2c2a2=b2−2acb2c2
Answered by A5T last updated on 22/Mar/24
aα^2 +bα+c=0; aβ^2 +bβ+c=0 ⇒?=(1/((−bα)^2 ))+(1/((−bβ)^2 ))=(1/b^2 )(((α^2 +β^2 )/((αβ)^2 )))=(((b^2 /a^2 )−((2c)/a))/(b^2 ((c^2 /a^2 )))) =((b^2 −2ac)/(b^2 c^2 ))
aα2+bα+c=0;aβ2+bβ+c=0⇒?=1(−bα)2+1(−bβ)2=1b2(α2+β2(αβ)2)=b2a2−2cab2(c2a2)=b2−2acb2c2

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