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 70617 by Rio Michael last updated on 06/Oct/19

given that  α and β are roots of the equation x^2 −5x + 4 =0   α>0 and β >0  find an equation whose roots are (√α) and (√β)     how do i find  (√(α )) + (√β)

$${given}\:{that}\:\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\:{the}\:{equation}\:{x}^{\mathrm{2}} −\mathrm{5}{x}\:+\:\mathrm{4}\:=\mathrm{0}\: \\ $$$$\alpha>\mathrm{0}\:{and}\:\beta\:>\mathrm{0} \\ $$$${find}\:{an}\:{equation}\:{whose}\:{roots}\:{are}\:\sqrt{\alpha}\:{and}\:\sqrt{\beta}\: \\ $$$$ \\ $$$${how}\:{do}\:{i}\:{find}\:\:\sqrt{\alpha\:}\:+\:\sqrt{\beta}\: \\ $$

Answered by mr W last updated on 06/Oct/19

αβ=4 ⇒(√α)(√β)=2  α+β=5   ⇒((√α))^2 +((√β))^2 =5  ⇒((√α))^2 +((√β))^2 +2(√( α))(√β)=5+2(√α)(√β)  ⇒((√α)+(√β))^2 =5+2×2=9  ⇒(√α)+(√β)=(√9)=3  the eqn. with roots (√α) and (√β) is  x^2 −3x+2=0

$$\alpha\beta=\mathrm{4}\:\Rightarrow\sqrt{\alpha}\sqrt{\beta}=\mathrm{2} \\ $$$$\alpha+\beta=\mathrm{5}\: \\ $$$$\Rightarrow\left(\sqrt{\alpha}\right)^{\mathrm{2}} +\left(\sqrt{\beta}\right)^{\mathrm{2}} =\mathrm{5} \\ $$$$\Rightarrow\left(\sqrt{\alpha}\right)^{\mathrm{2}} +\left(\sqrt{\beta}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\:\alpha}\sqrt{\beta}=\mathrm{5}+\mathrm{2}\sqrt{\alpha}\sqrt{\beta} \\ $$$$\Rightarrow\left(\sqrt{\alpha}+\sqrt{\beta}\right)^{\mathrm{2}} =\mathrm{5}+\mathrm{2}×\mathrm{2}=\mathrm{9} \\ $$$$\Rightarrow\sqrt{\alpha}+\sqrt{\beta}=\sqrt{\mathrm{9}}=\mathrm{3} \\ $$$${the}\:{eqn}.\:{with}\:{roots}\:\sqrt{\alpha}\:{and}\:\sqrt{\beta}\:{is} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$

Commented by Rio Michael last updated on 06/Oct/19

thank yiu so much mr W

$${thank}\:{yiu}\:{so}\:{much}\:{mr}\:{W} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com