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Question-66969




Question Number 66969 by mr W last updated on 21/Aug/19
Commented by mr W last updated on 21/Aug/19
Question from B.E.H.I. sir reposted.
$${Question}\:{from}\:{B}.{E}.{H}.{I}.\:{sir}\:{reposted}. \\ $$
Answered by mr W last updated on 21/Aug/19
y=((ax)/(x−b))  x^2 +((a^2 x^2 )/((x−b)^2 ))=ab  (x^2 −ab)(x−b)^2 +a^2 x^2 =0  (x^2 −ab)(x^2 −2bx+b^2 )+a^2 x^2 =0  x^4 −abx^2 −2bx^3 +2ab^2 x+b^2 x^2 −ab^3 +a^2 x^2 =0  x^4 −2bx^3 +(a^2 +b^2 −ab)x^2 +2ab^2 x−ab^3 =0  ((x/b))^4 −2((x/b))^3 +(1+(a^2 /b^2 )−(a/b))((x/b))^2 +2((a/b))((x/b))−(a/b)=0  let λ=(a/b), t=(x/b)  ⇒t^4 −2t^3 +(1+λ^2 −λ)t^2 +2λt−λ=0  can ajfour sir and MJS sir help to  solve this?
$${y}=\frac{{ax}}{{x}−{b}} \\ $$$${x}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {x}^{\mathrm{2}} }{\left({x}−{b}\right)^{\mathrm{2}} }={ab} \\ $$$$\left({x}^{\mathrm{2}} −{ab}\right)\left({x}−{b}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{ab}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{bx}+{b}^{\mathrm{2}} \right)+{a}^{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{abx}^{\mathrm{2}} −\mathrm{2}{bx}^{\mathrm{3}} +\mathrm{2}{ab}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {x}^{\mathrm{2}} −{ab}^{\mathrm{3}} +{a}^{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{bx}^{\mathrm{3}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right){x}^{\mathrm{2}} +\mathrm{2}{ab}^{\mathrm{2}} {x}−{ab}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\frac{{x}}{{b}}\right)^{\mathrm{4}} −\mathrm{2}\left(\frac{{x}}{{b}}\right)^{\mathrm{3}} +\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{{a}}{{b}}\right)\left(\frac{{x}}{{b}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{a}}{{b}}\right)\left(\frac{{x}}{{b}}\right)−\frac{{a}}{{b}}=\mathrm{0} \\ $$$${let}\:\lambda=\frac{{a}}{{b}},\:{t}=\frac{{x}}{{b}} \\ $$$$\Rightarrow{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} +\left(\mathrm{1}+\lambda^{\mathrm{2}} −\lambda\right){t}^{\mathrm{2}} +\mathrm{2}\lambda{t}−\lambda=\mathrm{0} \\ $$$${can}\:{ajfour}\:{sir}\:{and}\:{MJS}\:{sir}\:{help}\:{to} \\ $$$${solve}\:{this}? \\ $$

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