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If-and-are-the-roots-of-ax-2-bx-c-0-If-1-and-1-are-the-roots-of-a-1-x-2-b-1-x-c-1-0-If-b-1-a-1-k-b-c-then-k-

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Question Number 202120 by MATHEMATICSAM last updated on 21/Dec/23
If α and β are the roots of ax^2 + bx + c = 0. If ((1 − α)/α) and ((1 − β)/β) are the roots of a_1 x^2 + b_1 x + c_1 = 0. If (b_1 /a_1 ) = k + (b/c) then k = ?
$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}.\:\mathrm{If}\:\:\frac{\mathrm{1}\:−\:\alpha}{\alpha}\:\mathrm{and}\:\frac{\mathrm{1}\:−\:\beta}{\beta}\:\mathrm{are} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{a}_{\mathrm{1}} {x}^{\mathrm{2}} \:+\:{b}_{\mathrm{1}} {x}\:+\:{c}_{\mathrm{1}} \:=\:\mathrm{0}.\:\mathrm{If} \\ $$$$\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} }\:=\:{k}\:+\:\frac{{b}}{{c}}\:\mathrm{then}\:{k}\:=\:? \\ $$
Answered by cortano12 last updated on 21/Dec/23
{ ((ax^2 +bx+c=0⇒α+β=−(b/a))),((a_1 x^2 +b_1 x+c_1 =0⇒(1/α)+(1/β)−2=−(b_1 /a_1 ))) :} ⇒ −(b/c) −2 = −(b_1 /a_1 ) ⇒(b/c) + 2 = (b_1 /a_1 ) = k+(b/c) ∴ k = 2
$$\:\begin{cases}{{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\Rightarrow\alpha+\beta=−\frac{{b}}{{a}}}\\{{a}_{\mathrm{1}} {x}^{\mathrm{2}} +{b}_{\mathrm{1}} {x}+{c}_{\mathrm{1}} =\mathrm{0}\Rightarrow\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}−\mathrm{2}=−\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} }}\end{cases} \\ $$$$\:\Rightarrow\:−\frac{{b}}{{c}}\:−\mathrm{2}\:=\:−\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} } \\ $$$$\:\Rightarrow\frac{{b}}{{c}}\:+\:\mathrm{2}\:=\:\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} }\:=\:{k}+\frac{{b}}{{c}} \\ $$$$\:\therefore\:{k}\:=\:\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Dec/23
(b_1 /a_1 ) = k + (b/c) k=(b_1 /a_1 )−(b/c) =−(−(b_1 /a_1 ))+(−(b/a)÷(c/a)) =−(((1 − α)/α) + ((1 − β)/β))+(α+β)÷αβ =−(((β−αβ+α−αβ)/(αβ)))+((α+β)/(αβ)) =((−(α+β)+2αβ+(α+β))/(αβ))=2
$$\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} }\:=\:{k}\:+\:\frac{{b}}{{c}} \\ $$$${k}=\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} }−\frac{{b}}{{c}} \\ $$$$\:\:=−\left(−\frac{{b}_{\mathrm{1}} }{{a}_{\mathrm{1}} }\right)+\left(−\frac{{b}}{{a}}\boldsymbol{\div}\frac{{c}}{{a}}\right) \\ $$$$\:=−\left(\frac{\mathrm{1}\:−\:\alpha}{\alpha}\:+\:\frac{\mathrm{1}\:−\:\beta}{\beta}\right)+\left(\alpha+\beta\right)\boldsymbol{\div}\alpha\beta \\ $$$$\:=−\left(\frac{\beta−\alpha\beta+\alpha−\alpha\beta}{\alpha\beta}\right)+\frac{\alpha+\beta}{\alpha\beta} \\ $$$$\:\:=\frac{−\cancel{\left(\alpha+\beta\right)}+\mathrm{2}\alpha\beta+\cancel{\left(\alpha+\beta\right)}}{\alpha\beta}=\mathrm{2} \\ $$

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