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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 )) = ?
$$\mathrm{If}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\alpha\:\mathrm{and}\: \\ $$$$\beta\:\mathrm{then}\:\frac{\mathrm{1}}{\left({a}\alpha^{\mathrm{2}} \:+\:{c}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left({a}\beta^{\mathrm{2}} \:+\:{c}\right)^{\mathrm{2}} }\:=\:? \\ $$
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 ))
$$\mathrm{Another}\:\mathrm{Way} \\ $$$$\:\frac{\mathrm{1}}{\left({a}\alpha^{\mathrm{2}} \:+\:{c}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left({a}\beta^{\mathrm{2}} \:+\:{c}\right)^{\mathrm{2}} }\: \\ $$$$\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(\alpha^{\mathrm{2}} \:+\:\frac{{c}}{{a}}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(\beta^{\mathrm{2}} \:+\:\frac{{c}}{{a}}\right)^{\mathrm{2}} }\: \\ $$$$\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(\alpha^{\mathrm{2}} \:+\:\alpha\beta\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(\beta^{\mathrm{2}} \:+\:\alpha\beta\right)^{\mathrm{2}} }\: \\ $$$$\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \alpha^{\mathrm{2}} \left(\alpha\:+\:\beta\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \beta^{\mathrm{2}} \left(\beta\:+\:\alpha\right)^{\mathrm{2}} }\: \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(\alpha+\beta\right)^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(\alpha+\beta\right)^{\mathrm{2}} }\centerdot\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\left(\alpha\beta\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(\alpha+\beta\right)^{\mathrm{2}} }\centerdot\frac{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta}{\left(\alpha\beta\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left(−\frac{{b}}{{a}}\right)^{\mathrm{2}} }\centerdot\frac{\left(−\frac{{b}}{{a}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{c}}{{a}}\right)}{\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\centerdot\frac{\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}}{\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\centerdot\frac{\frac{{b}^{\mathrm{2}} −\mathrm{2}{ac}}{{a}^{\mathrm{2}} }}{\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }} \\ $$$$=\frac{{b}^{\mathrm{2}} −\mathrm{2}{ac}}{{b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$
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 ))
$${a}\alpha^{\mathrm{2}} +{b}\alpha+{c}=\mathrm{0};\:{a}\beta^{\mathrm{2}} +{b}\beta+{c}=\mathrm{0} \\ $$$$\Rightarrow?=\frac{\mathrm{1}}{\left(−{b}\alpha\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(−{b}\beta\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\left(\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\left(\alpha\beta\right)^{\mathrm{2}} }\right)=\frac{\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}}{{b}^{\mathrm{2}} \left(\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)} \\ $$$$=\frac{{b}^{\mathrm{2}} −\mathrm{2}{ac}}{{b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$

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