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

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