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Question Number 138766 by nadovic last updated on 18/Apr/21
If α and β are the roots of the equation 3x^2 +x+2=0, find the equation whose roots are (1/α^2 ) and (1/β^2 ) and show that 27α^4 =11α+10.
$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +{x}+\mathrm{2}=\mathrm{0},\:\mathrm{find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{whose} \\ $$$$\mathrm{roots}\:\mathrm{are}\:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\:\mathrm{and}\:\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:\:\mathrm{and}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{27}\alpha^{\mathrm{4}} =\mathrm{11}\alpha+\mathrm{10}. \\ $$
Commented by MJS_new last updated on 18/Apr/21
x^2 +px+q=0 with roots α, β ⇒ x^2 +((2q−p^2 )/q^2 )x+(1/q^2 )=0 with roots (1/α^2 ), (1/β^2 )
$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\:\mathrm{with}\:\mathrm{roots}\:\alpha,\:\beta \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{2}{q}−{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }{x}+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }=\mathrm{0}\:\mathrm{with}\:\mathrm{roots}\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} },\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} } \\ $$
Answered by bramlexs22 last updated on 18/Apr/21
3x^2 +x+2=0→ { (α),(β) :} ⇒3((√x))^2 +(√x)+2=0 ⇒(√x) = −2−3x ⇒x=4+12x+9x^2 ⇒9x^2 +11x+4=0→ { (α^2 ),(β^2 ) :} and 4x^2 +11x+9=0→ { ((1/α^2 )),((1/β^2 )) :}
$$\:\mathrm{3}{x}^{\mathrm{2}} +{x}+\mathrm{2}=\mathrm{0}\rightarrow\begin{cases}{\alpha}\\{\beta}\end{cases} \\ $$$$\Rightarrow\mathrm{3}\left(\sqrt{{x}}\right)^{\mathrm{2}} +\sqrt{{x}}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\sqrt{{x}}\:=\:−\mathrm{2}−\mathrm{3}{x}\: \\ $$$$\Rightarrow{x}=\mathrm{4}+\mathrm{12}{x}+\mathrm{9}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{4}=\mathrm{0}\rightarrow\begin{cases}{\alpha^{\mathrm{2}} }\\{\beta^{\mathrm{2}} }\end{cases} \\ $$$${and}\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{9}=\mathrm{0}\rightarrow\begin{cases}{\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }}\\{\frac{\mathrm{1}}{\beta^{\mathrm{2}} }}\end{cases} \\ $$
Commented by Rasheed.Sindhi last updated on 18/Apr/21
⇒x=4+12x+3x^2 ⇒^? 3x^2 +11x+4=0→ { (α^2 ),(β^2 ) :}
$$\Rightarrow{x}=\mathrm{4}+\mathrm{12}{x}+\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\overset{?} {\Rightarrow}\mathrm{3}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{4}=\mathrm{0}\rightarrow\begin{cases}{\alpha^{\mathrm{2}} }\\{\beta^{\mathrm{2}} }\end{cases} \\ $$$$ \\ $$
Commented by bramlexs22 last updated on 18/Apr/21
typo
$${typo} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Apr/21
α+β=−1/3 , αβ=2/3 (1/α^2 )+(1/β^2 )=((α^2 +β^2 )/(α^2 β^2 ))=(((α+β)^2 −2αβ)/((αβ)^2 )) =(((−1/3)^2 −2(2/3))/((2/3)^2 ))=(((1/9)−(4/3))/(4/9)) =((1−12)/4)=−((11)/4) (1/α^2 ).(1/β^2 )=(1/((αβ)^2 ))=(1/((2/3)^2 ))=(9/4) x^2 −((1/α^2 )+(1/β^2 ))x+(1/α^2 ).(1/β^2 )=0 x^2 −(−((11)/4))x+(9/4)=0 4x^2 +11x+9=0
$$\alpha+\beta=−\mathrm{1}/\mathrm{3}\:\:\:,\:\:\:\:\alpha\beta=\mathrm{2}/\mathrm{3} \\ $$$$\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} }=\frac{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta}{\left(\alpha\beta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\frac{\left(−\mathrm{1}/\mathrm{3}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}/\mathrm{3}\right)}{\left(\mathrm{2}/\mathrm{3}\right)^{\mathrm{2}} }=\frac{\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{4}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{9}}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}−\mathrm{12}}{\mathrm{4}}=−\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }.\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\alpha\beta\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\mathrm{2}/\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\right){x}+\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }.\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\left(−\frac{\mathrm{11}}{\mathrm{4}}\right){x}+\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{9}=\mathrm{0} \\ $$$$ \\ $$
Answered by physicstutes last updated on 18/Apr/21
α + β = −(b/a) = −(1/3) αβ = (c/a) = (2/3) sum of roots: (1/α^2 ) + (1/β^2 ) = (((α+β)^2 −2αβ)/((αβ)^2 )) = ((((1/9))−(4/3))/(4/9)) = ((−11)/4) product of roots: ((1/α^2 ))((1/β^2 )) = (1/((αβ)^2 )) = (9/4) new equation: x^2 +((11)/4)x + (9/4) = 0 ⇒ 4x^2 +11x + 9 = 0
$$\alpha\:+\:\beta\:=\:−\frac{{b}}{{a}}\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\alpha\beta\:=\:\frac{{c}}{{a}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{roots}:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:=\:\frac{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta}{\left(\alpha\beta\right)^{\mathrm{2}} }\:=\:\frac{\left(\frac{\mathrm{1}}{\mathrm{9}}\right)−\frac{\mathrm{4}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{9}}}\:=\:\frac{−\mathrm{11}}{\mathrm{4}} \\ $$$$\mathrm{product}\:\mathrm{of}\:\mathrm{roots}:\:\left(\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\right)\:=\:\frac{\mathrm{1}}{\left(\alpha\beta\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{new}\:\mathrm{equation}:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}}{x}\:+\:\frac{\mathrm{9}}{\mathrm{4}}\:=\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{11}{x}\:+\:\mathrm{9}\:=\:\mathrm{0} \\ $$

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