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Question Number 96189 by bemath last updated on 30/May/20

find a common roots from the two quadratic eq 24x^2 +(p+4)x−1=0 and 6x^2 +11x+p+2=0

$$\mathrm{find}\:\mathrm{a}\:\mathrm{common}\:\mathrm{roots}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{quadratic}\:\mathrm{eq} \\ $$$$\mathrm{24x}^{\mathrm{2}} +\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{6x}^{\mathrm{2}} +\mathrm{11x}+\mathrm{p}+\mathrm{2}=\mathrm{0} \\ $$

Answered by john santu last updated on 30/May/20

by Wildberger′s theorem if two quadratic eq : x^2 +Ax+B=0 and x^2 +Cx+D=0 have a common roots it′s ⇒ x = ((D−B)/(A−C)) (i)x^2 +(((p+4)/(24)))x−(1/(24)) = 0 (ii) x^2 +(((11)/6))x+(((p+2)/6))=0 so the common roots is x = ((((p+2)/6)+(1/(24)))/((((p+4)/(24)))−((11)/6))) = ((4p+9)/(p+4−44)) x = ((4p+9)/(p−40))

$$\mathrm{by}\:\mathrm{Wildberger}'\mathrm{s}\:\mathrm{theorem} \\ $$$$\mathrm{if}\:\mathrm{two}\:\mathrm{quadratic}\:\mathrm{eq}\::\:\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}+\mathrm{B}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{x}^{\mathrm{2}} +\mathrm{Cx}+\mathrm{D}=\mathrm{0}\:\mathrm{have}\:\mathrm{a}\:\mathrm{common} \\ $$$$\mathrm{roots}\:\mathrm{it}'\mathrm{s}\:\Rightarrow\:{x}\:=\:\frac{{D}−\mathrm{B}}{\mathrm{A}−\mathrm{C}} \\ $$$$\left(\mathrm{i}\right){x}^{\mathrm{2}} +\left(\frac{{p}+\mathrm{4}}{\mathrm{24}}\right){x}−\frac{\mathrm{1}}{\mathrm{24}}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{ii}\right)\:{x}^{\mathrm{2}} +\left(\frac{\mathrm{11}}{\mathrm{6}}\right){x}+\left(\frac{{p}+\mathrm{2}}{\mathrm{6}}\right)=\mathrm{0} \\ $$$${so}\:{the}\:{common}\:{roots}\: \\ $$$${is}\:{x}\:=\:\frac{\frac{{p}+\mathrm{2}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}}{\left(\frac{{p}+\mathrm{4}}{\mathrm{24}}\right)−\frac{\mathrm{11}}{\mathrm{6}}}\:=\:\frac{\mathrm{4}{p}+\mathrm{9}}{{p}+\mathrm{4}−\mathrm{44}} \\ $$$${x}\:=\:\frac{\mathrm{4}{p}+\mathrm{9}}{{p}−\mathrm{40}}\: \\ $$

Commented by bemath last updated on 30/May/20

wow ... I just found out the theorem mister

Commented by mr W last updated on 30/May/20

only for certain values of p have both equations a common root.

$${only}\:{for}\:{certain}\:{values}\:{of}\:{p}\:{have}\:{both} \\ $$$${equations}\:{a}\:{common}\:{root}. \\ $$

Commented by john santu last updated on 30/May/20

Yes, right

Answered by mr W last updated on 30/May/20

let′s say eqn. 1 has roots: α, β eqn. 2 has roots: β, γ α+β=−((p+4)/(24)) αβ=−(1/(24)) β+γ=−((11)/6) βγ=((p+2)/6) ⇒(((p+4)/(24))+β)β=(1/(24)) ...(i) ⇒(((11)/6)+β)β=−((p+2)/6) ...(ii) from (ii): ((p+4)/(24))=(1/(12))−(1/4)(((11)/6)+β)β put this into (i): 6β^3 −13β^2 −2β+1=0 ⇒(3β+1)(2β^2 −5β+1)=0 ⇒3β+1=0 ⇒β=−(1/3) ⇒2β^2 −5β+1=0 ⇒β=((5±(√(17)))/4) i.e. the common root can be −(1/3), ((5−(√(17)))/4), ((5+(√(17)))/4) p=−2−6(((11)/6)+β)β ⇒p=1 ⇒p=−((63−13(√(17)))/2)≈−4.7 ⇒p=−((63+13(√(17)))/2)≈−58.3

$${let}'{s}\:{say}\:{eqn}.\:\mathrm{1}\:{has}\:{roots}:\:\alpha,\:\beta \\ $$$${eqn}.\:\mathrm{2}\:{has}\:{roots}:\:\beta,\:\gamma \\ $$$$\alpha+\beta=−\frac{{p}+\mathrm{4}}{\mathrm{24}} \\ $$$$\alpha\beta=−\frac{\mathrm{1}}{\mathrm{24}} \\ $$$$\beta+\gamma=−\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$\beta\gamma=\frac{{p}+\mathrm{2}}{\mathrm{6}} \\ $$$$\Rightarrow\left(\frac{{p}+\mathrm{4}}{\mathrm{24}}+\beta\right)\beta=\frac{\mathrm{1}}{\mathrm{24}}\:\:\:...\left({i}\right) \\ $$$$\Rightarrow\left(\frac{\mathrm{11}}{\mathrm{6}}+\beta\right)\beta=−\frac{{p}+\mathrm{2}}{\mathrm{6}}\:\:\:...\left({ii}\right) \\ $$$${from}\:\left({ii}\right): \\ $$$$\frac{{p}+\mathrm{4}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{11}}{\mathrm{6}}+\beta\right)\beta \\ $$$${put}\:{this}\:{into}\:\left({i}\right): \\ $$$$\mathrm{6}\beta^{\mathrm{3}} −\mathrm{13}\beta^{\mathrm{2}} −\mathrm{2}\beta+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3}\beta+\mathrm{1}\right)\left(\mathrm{2}\beta^{\mathrm{2}} −\mathrm{5}\beta+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}\beta+\mathrm{1}=\mathrm{0}\:\Rightarrow\beta=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}\beta^{\mathrm{2}} −\mathrm{5}\beta+\mathrm{1}=\mathrm{0}\:\Rightarrow\beta=\frac{\mathrm{5}\pm\sqrt{\mathrm{17}}}{\mathrm{4}} \\ $$$$ \\ $$$${i}.{e}.\:{the}\:{common}\:{root}\:{can}\:{be} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{5}−\sqrt{\mathrm{17}}}{\mathrm{4}},\:\frac{\mathrm{5}+\sqrt{\mathrm{17}}}{\mathrm{4}} \\ $$$$ \\ $$$${p}=−\mathrm{2}−\mathrm{6}\left(\frac{\mathrm{11}}{\mathrm{6}}+\beta\right)\beta \\ $$$$\Rightarrow{p}=\mathrm{1} \\ $$$$\Rightarrow{p}=−\frac{\mathrm{63}−\mathrm{13}\sqrt{\mathrm{17}}}{\mathrm{2}}\approx−\mathrm{4}.\mathrm{7} \\ $$$$\Rightarrow{p}=−\frac{\mathrm{63}+\mathrm{13}\sqrt{\mathrm{17}}}{\mathrm{2}}\approx−\mathrm{58}.\mathrm{3} \\ $$

Answered by 1549442205 last updated on 30/May/20

Denoting x_0 common root of two equations.Then { (( 24x_0 ^2 +(p+4)x_0 −1=0(1))),((6x_0 ^2 +11x_0 +p+2=0 (2))) :} ⇔ { ((24x_0 ^2 +(p+4)x_0 −1=0)),((24x_0 +44x_0 +4p+8=0)) :} ⇒(p+4)x_0 −1=44x_0 +4p+8 ⇒(p−40)x_0 =4p+9⇒x_0 =((4p+9)/(p−40)).Replace into (2) ⇒6(((4p+9)/(p−40)))^2 +11(((4p+9)/(p−40)))+p+2=0 ⇔6(16p^2 +72p+81)+11(4p^2 −151p−360) +(p+2)(p^2 −80p+1600)=0⇔ 140p^2 −1229p−3474+p^3 −78p^2 +1440p +3200=0⇔p^3 +62p^2 +211p−274=0 ⇔(p−1)(p^2 +63p+274)=0 a/p=1⇒x_0 =((−1)/3) b/p^2 +63p+274=0⇔p=((−63±13(√(17)))/2) ⇒x_0 ∈{((5−(√(17)))/4);(((√(17))+5)/4)} Thus,two given equations have three common

$$\mathrm{Denoting}\:\mathrm{x}_{\mathrm{0}} \:\mathrm{common}\:\mathrm{root}\:\mathrm{of}\:\mathrm{two}\:\mathrm{equations}.\mathrm{Then} \\ $$$$\begin{cases}{\:\mathrm{24x}_{\mathrm{0}} ^{\mathrm{2}} \:+\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}_{\mathrm{0}} −\mathrm{1}=\mathrm{0}\left(\mathrm{1}\right)}\\{\mathrm{6x}_{\mathrm{0}} ^{\mathrm{2}} \:+\mathrm{11x}_{\mathrm{0}} +\mathrm{p}+\mathrm{2}=\mathrm{0}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{24x}_{\mathrm{0}} ^{\mathrm{2}} \:+\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}_{\mathrm{0}} −\mathrm{1}=\mathrm{0}}\\{\mathrm{24x}_{\mathrm{0}} +\mathrm{44x}_{\mathrm{0}} +\mathrm{4p}+\mathrm{8}=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}_{\mathrm{0}} −\mathrm{1}=\mathrm{44x}_{\mathrm{0}} +\mathrm{4p}+\mathrm{8} \\ $$$$\Rightarrow\left(\mathrm{p}−\mathrm{40}\right)\mathrm{x}_{\mathrm{0}} =\mathrm{4p}+\mathrm{9}\Rightarrow\mathrm{x}_{\mathrm{0}} =\frac{\mathrm{4p}+\mathrm{9}}{\mathrm{p}−\mathrm{40}}.\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{6}\left(\frac{\mathrm{4p}+\mathrm{9}}{\mathrm{p}−\mathrm{40}}\right)^{\mathrm{2}} +\mathrm{11}\left(\frac{\mathrm{4p}+\mathrm{9}}{\mathrm{p}−\mathrm{40}}\right)+\mathrm{p}+\mathrm{2}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{6}\left(\mathrm{16p}^{\mathrm{2}} +\mathrm{72p}+\mathrm{81}\right)+\mathrm{11}\left(\mathrm{4p}^{\mathrm{2}} −\mathrm{151p}−\mathrm{360}\right) \\ $$$$+\left(\mathrm{p}+\mathrm{2}\right)\left(\mathrm{p}^{\mathrm{2}} −\mathrm{80p}+\mathrm{1600}\right)=\mathrm{0}\Leftrightarrow \\ $$$$\mathrm{140p}^{\mathrm{2}} −\mathrm{1229p}−\mathrm{3474}+\mathrm{p}^{\mathrm{3}} −\mathrm{78p}^{\mathrm{2}} +\mathrm{1440p} \\ $$$$+\mathrm{3200}=\mathrm{0}\Leftrightarrow\mathrm{p}^{\mathrm{3}} +\mathrm{62p}^{\mathrm{2}} +\mathrm{211p}−\mathrm{274}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{p}−\mathrm{1}\right)\left(\mathrm{p}^{\mathrm{2}} +\mathrm{63p}+\mathrm{274}\right)=\mathrm{0} \\ $$$$\mathrm{a}/\mathrm{p}=\mathrm{1}\Rightarrow\mathrm{x}_{\mathrm{0}} =\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{b}/\mathrm{p}^{\mathrm{2}} +\mathrm{63p}+\mathrm{274}=\mathrm{0}\Leftrightarrow\mathrm{p}=\frac{−\mathrm{63}\pm\mathrm{13}\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{0}} \in\left\{\frac{\mathrm{5}−\sqrt{\mathrm{17}}}{\mathrm{4}};\frac{\sqrt{\mathrm{17}}+\mathrm{5}}{\mathrm{4}}\right\} \\ $$$$\mathrm{Thus},\mathrm{two}\:\mathrm{given}\:\mathrm{equations}\:\mathrm{have}\:\mathrm{three} \\ $$$$\mathrm{common} \\ $$

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