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and-are-2-roots-of-eq-ax-2-bx-c-0-with-conditions-2-b-2-a-find-c-in-terms-of-a-and-b-

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Question Number 55039 by behi83417@gmail.com last updated on 16/Feb/19
α and β,are 2 roots of eq: ax^2 +bx+c=0 with conditions: { ((α^2 =β+b)),((β^2 =α+a)) :} find: c in terms of: a and b.
$$\alpha\:{and}\:\beta,{are}\:\mathrm{2}\:{roots}\:{of}\:{eq}: \\ $$$$\:\:\:\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{with}\:{conditions}: \\ $$$$\:\:\:\:\begin{cases}{\alpha^{\mathrm{2}} =\beta+{b}}\\{\beta^{\mathrm{2}} =\alpha+{a}}\end{cases} \\ $$$${find}:\:\:\boldsymbol{{c}}\:{in}\:{terms}\:{of}:\:\boldsymbol{{a}}\:\:{and}\:\:\boldsymbol{{b}}. \\ $$
Commented by mr W last updated on 17/Feb/19
i think the question has no solution, because there are too many conditions. α and β should be roots of ax^2 +bx+c=0, but they are already given by eqn. system { ((α^2 =β+b)),((β^2 =α+a)) :} this is not possible with given a and b as shown below. α+β=−(b/a) αβ=(c/a) on one side: α^2 +β^2 =α+β+a+b (α+β)^2 −2αβ=α+β+a+b (b^2 /a^2 )−((2c)/a)=−(b/a)+(a+b) b^2 −2ac=−ab+(a+b)a^2 ⇒c=(((b−a^2 )(a+b))/(2a)) ...(i) but on the other side: α^2 −β^2 =β−α+a−b (α−β)(α+β)=β−α+a−b (α−β)(1+α+β)=a−b (α−β)(1−(b/a))=a−b α−β=a α^2 +β^2 −2αβ=a^2 (b^2 /a^2 )−((2c)/a)=a^2 ((2c)/a)=(b^2 /a^2 )−a^2 =(((b+a^2 )(b−a^2 ))/a^2 ) ⇒c=(((b−a^2 )(a^2 +b))/(2a)) ...(ii) (i) and (ii) are contradiction.
$${i}\:{think}\:{the}\:{question}\:{has}\:{no}\:{solution}, \\ $$$${because}\:{there}\:{are}\:{too}\:{many}\:{conditions}. \\ $$$$ \\ $$$$\alpha\:{and}\:\beta\:{should}\:{be}\:{roots}\:{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}, \\ $$$${but}\:{they}\:{are}\:{already}\:{given}\:{by}\:{eqn}.\:{system} \\ $$$$\:\:\:\:\begin{cases}{\alpha^{\mathrm{2}} =\beta+{b}}\\{\beta^{\mathrm{2}} =\alpha+{a}}\end{cases} \\ $$$${this}\:{is}\:{not}\:{possible}\:{with}\:{given}\:{a}\:{and}\:{b}\:{as}\:{shown}\:{below}. \\ $$$$ \\ $$$$\alpha+\beta=−\frac{{b}}{{a}} \\ $$$$\alpha\beta=\frac{{c}}{{a}} \\ $$$${on}\:{one}\:{side}: \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\alpha+\beta+{a}+{b} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta=\alpha+\beta+{a}+{b} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}=−\frac{{b}}{{a}}+\left({a}+{b}\right) \\ $$$${b}^{\mathrm{2}} −\mathrm{2}{ac}=−{ab}+\left({a}+{b}\right){a}^{\mathrm{2}} \\ $$$$\Rightarrow{c}=\frac{\left({b}−{a}^{\mathrm{2}} \right)\left({a}+{b}\right)}{\mathrm{2}{a}}\:\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${but}\:{on}\:{the}\:{other}\:{side}: \\ $$$$\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} =\beta−\alpha+{a}−{b} \\ $$$$\left(\alpha−\beta\right)\left(\alpha+\beta\right)=\beta−\alpha+{a}−{b} \\ $$$$\left(\alpha−\beta\right)\left(\mathrm{1}+\alpha+\beta\right)={a}−{b} \\ $$$$\left(\alpha−\beta\right)\left(\mathrm{1}−\frac{{b}}{{a}}\right)={a}−{b} \\ $$$$\alpha−\beta={a} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\mathrm{2}\alpha\beta={a}^{\mathrm{2}} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}={a}^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{c}}{{a}}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−{a}^{\mathrm{2}} =\frac{\left({b}+{a}^{\mathrm{2}} \right)\left({b}−{a}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{c}=\frac{\left({b}−{a}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}\right)}{\mathrm{2}{a}}\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)\:{and}\:\left({ii}\right)\:{are}\:{contradiction}. \\ $$
Commented by kaivan.ahmadi last updated on 17/Feb/19
hi mr w please check the proof again α^2 −β^2 =(β−α)+(b−a)⇒ (α−β)(1+(α+β))=b−a
$${hi}\:{mr}\:{w} \\ $$$${please}\:{check}\:{the}\:{proof}\:{again} \\ $$$$\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} =\left(\beta−\alpha\right)+\left({b}−{a}\right)\Rightarrow \\ $$$$\left(\alpha−\beta\right)\left(\mathrm{1}+\left(\alpha+\beta\right)\right)={b}−{a} \\ $$$$ \\ $$
Commented by mr W last updated on 17/Feb/19
thank you sir! you′re right. i′ve fixed. please check again.
$${thank}\:{you}\:{sir}!\:{you}'{re}\:{right}.\:{i}'{ve}\:{fixed}. \\ $$$${please}\:{check}\:{again}. \\ $$
Answered by kaivan.ahmadi last updated on 16/Feb/19
{ ((α^2 −β=b)),((β^2 −α=a)) :}⇒(α^2 +β^2 )−(α+β)=a+b⇒ (α+β)^2 −2αβ−(α+β)=a+b⇒ (((−b)/a))^2 −((2c)/a)−(((−b)/a))=a+b⇒ ((2c)/a)=(b^2 /a^2 )+(b/a)−a−b⇒c=(b^2 /(2a))+(b/2)−(a^2 /2)−((ab)/2)= ((b^2 +ab−a^3 −a^2 b)/(2a))
$$ \\ $$$$\begin{cases}{\alpha^{\mathrm{2}} −\beta={b}}\\{\beta^{\mathrm{2}} −\alpha={a}}\end{cases}\Rightarrow\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)−\left(\alpha+\beta\right)={a}+{b}\Rightarrow \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta−\left(\alpha+\beta\right)={a}+{b}\Rightarrow \\ $$$$\left(\frac{−{b}}{{a}}\right)^{\mathrm{2}} −\frac{\mathrm{2}{c}}{{a}}−\left(\frac{−{b}}{{a}}\right)={a}+{b}\Rightarrow \\ $$$$\frac{\mathrm{2}{c}}{{a}}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{b}}{{a}}−{a}−{b}\Rightarrow{c}=\frac{{b}^{\mathrm{2}} }{\mathrm{2}{a}}+\frac{{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{{ab}}{\mathrm{2}}= \\ $$$$\frac{{b}^{\mathrm{2}} +{ab}−{a}^{\mathrm{3}} −{a}^{\mathrm{2}} {b}}{\mathrm{2}{a}} \\ $$

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