Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\alpha\:+\:\beta\:=\:−\frac{{b}}{{a}}\:=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\alpha\beta\:=\:\frac{{c}}{{a}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{roots}:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:=\:\frac{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta}{\left(\alpha\beta\right)^{\mathrm{2}} }\:=\:\frac{\left(\frac{\mathrm{1}}{\mathrm{9}}\right)−\frac{\mathrm{4}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{9}}}\:=\:\frac{−\mathrm{11}}{\mathrm{4}} \\ $$$$\mathrm{product}\:\mathrm{of}\:\mathrm{roots}:\:\left(\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\right)\:=\:\frac{\mathrm{1}}{\left(\alpha\beta\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{new}\:\mathrm{equation}:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}}{x}\:+\:\frac{\mathrm{9}}{\mathrm{4}}\:=\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{11}{x}\:+\:\mathrm{9}\:=\:\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com