Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 33139 by Rio Mike last updated on 11/Apr/18

α and β are roots of ax^2 +bx +c=0 show that α+β= ((−b)/a) and αβ= (c/a) hence form an equation whose sum of roots and product of roots are respectively −(1/2) and 2.

$$\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\: \\ $$$$\:{ax}^{\mathrm{2}} +{bx}\:+{c}=\mathrm{0} \\ $$$${show}\:{that}\:\alpha+\beta=\:\frac{−{b}}{{a}}\:{and}\:\alpha\beta=\:\frac{{c}}{{a}} \\ $$$${hence}\:{form}\:{an}\:{equation}\:{whose} \\ $$$${sum}\:{of}\:{roots}\:{and}\:{product}\:{of}\:{roots} \\ $$$${are}\:{respectively}\: \\ $$$$\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\mathrm{2}. \\ $$

Answered by MJS last updated on 11/Apr/18

ax^2 +bx+c=0 x^2 +(b/a)x+(c/a)=0 (x−α)(x−β)=0 x^2 −(α+β)x+αβ=0 ⇒ α+β=−(b/a); αβ=(c/a) x^2 +(1/2)x+2=0 x=−(1/4)±(√((1/(16))−2))=−(1/4)±((√(31))/4)i α=−(1/4)−((√(31))/4)i β=−(1/4)+((√(31))/4)i α+β=−(1/2) αβ=(−(1/4)−((√(31))/4)i)(−(1/4)+((√(31))/4)i)= =(1/(16))−((31)/(16))i^2 =(1/(16))+((31)/(16))=2

$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}=\mathrm{0} \\ $$$$ \\ $$$$\left({x}−\alpha\right)\left({x}−\beta\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\left(\alpha+\beta\right){x}+\alpha\beta=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:\alpha+\beta=−\frac{{b}}{{a}};\:\alpha\beta=\frac{{c}}{{a}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{2}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{4}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{16}}−\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}{i} \\ $$$$\alpha=−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}{i} \\ $$$$\beta=−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}{i} \\ $$$$\alpha+\beta=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\alpha\beta=\left(−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}{i}\right)\left(−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}{i}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{31}}{\mathrm{16}}{i}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{31}}{\mathrm{16}}=\mathrm{2} \\ $$

Commented by Rio Mike last updated on 11/Apr/18

Generally the new equation is 2x^2 +x +4=0 since Sum of roots α+β= −(1/2) αβ=2 New equation = X^2 −(Sumroots)X +product of roots x^2 −−(1/2)x + 2 2x^2 +x+4=0

$${Generally}\:{the}\:{new}\:{equation}\:{is} \\ $$$$\:\:\:\mathrm{2}{x}^{\mathrm{2}} +{x}\:+\mathrm{4}=\mathrm{0} \\ $$$${since}\: \\ $$$${Sum}\:{of}\:{roots}\:\alpha+\beta=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\alpha\beta=\mathrm{2} \\ $$$${New}\:{equation}\:=\:{X}^{\mathrm{2}} −\left({Sumroots}\right){X}\:+{product}\:{of}\:{roots} \\ $$$${x}^{\mathrm{2}} −−\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\:\mathrm{2} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{4}=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com