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Question Number 155650 by mathdanisur last updated on 03/Oct/21
Solve for real numbers: (√((x-a)/(x-b))) + (a/x) = (√((x-b)/(x-a))) + (b/x) a;b∈R and a≠b
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}-\mathrm{b}}}\:+\:\frac{\mathrm{a}}{\mathrm{x}}\:=\:\sqrt{\frac{\mathrm{x}-\mathrm{b}}{\mathrm{x}-\mathrm{a}}}\:+\:\frac{\mathrm{b}}{\mathrm{x}} \\ $$$$\mathrm{a};\mathrm{b}\in\mathbb{R}\:\:\mathrm{and}\:\:\mathrm{a}\neq\mathrm{b} \\ $$
Answered by som(math1967) last updated on 03/Oct/21
(√((x−a)/(x−b)))−(√((x−b)/(x−a)))=((b−a)/x) ((x−a−x+b)/( (√(x^2 −(a+b)x+ab))))=((b−a)/x) (1/( (√(x^2 −(a+b)x+ab))))=(1/x) [∵a≠b] x^2 −(a+b)x+ab=x^2 [squaring both side] x=((ab)/(a+b)) ans
$$\sqrt{\frac{{x}−{a}}{{x}−{b}}}−\sqrt{\frac{{x}−{b}}{{x}−{a}}}=\frac{{b}−{a}}{{x}} \\ $$$$\frac{{x}−{a}−{x}+{b}}{\:\sqrt{{x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}}}=\frac{{b}−{a}}{{x}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}}}=\frac{\mathrm{1}}{{x}}\:\:\left[\because{a}\neq{b}\right] \\ $$$${x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}={x}^{\mathrm{2}} \:\:\left[{squaring}\:{both}\:{side}\right] \\ $$$${x}=\frac{{ab}}{{a}+{b}}\:{ans} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 03/Oct/21
You attack a few questions but your answers are always perfect! :)
$${You}\:{attack}\:{a}\:{few}\:{questions}\:{but} \\ $$$$\left.{your}\:{answers}\:{are}\:{always}\:{perfect}!\::\right) \\ $$
Commented by som(math1967) last updated on 03/Oct/21
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$
Answered by ajfour last updated on 03/Oct/21
((x−a)/(x−b))=t^2 ⇒ x=((bt^2 −a)/(t^2 −1)) t−(1/t)=(((b−a)(t^2 −1))/(bt^2 −a)) as t≠1 ⇒ bt^2 −a=(b−a)t x=b+((b−a)/(t^2 −1)) =b+((b(b−a))/((b−a)t−(b−a))) x=b+(b/(t−1)) t=(((b−a))/(2b))±(√((((b−a)^2 )/(4b^2 ))+(a/b))) t=((b−a±∣a+b∣)/(2b)) x=b+((2b^2 )/(−a−b±∣a+b∣)) rejecting +ve sign ⇒ x=b−(b^2 /(a+b)) x=((ab)/(a+b)) .
$$\frac{{x}−{a}}{{x}−{b}}={t}^{\mathrm{2}} \:\:\Rightarrow\:\:{x}=\frac{{bt}^{\mathrm{2}} −{a}}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$${t}−\frac{\mathrm{1}}{{t}}=\frac{\left({b}−{a}\right)\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{{bt}^{\mathrm{2}} −{a}} \\ $$$${as}\:\:{t}\neq\mathrm{1}\:\:\Rightarrow\:\:{bt}^{\mathrm{2}} −{a}=\left({b}−{a}\right){t} \\ $$$${x}={b}+\frac{{b}−{a}}{{t}^{\mathrm{2}} −\mathrm{1}}\:\:={b}+\frac{{b}\left({b}−{a}\right)}{\left({b}−{a}\right){t}−\left({b}−{a}\right)} \\ $$$${x}={b}+\frac{{b}}{{t}−\mathrm{1}} \\ $$$${t}=\frac{\left({b}−{a}\right)}{\mathrm{2}{b}}\pm\sqrt{\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }+\frac{{a}}{{b}}} \\ $$$${t}=\frac{{b}−{a}\pm\mid{a}+{b}\mid}{\mathrm{2}{b}} \\ $$$${x}={b}+\frac{\mathrm{2}{b}^{\mathrm{2}} }{−{a}−{b}\pm\mid{a}+{b}\mid} \\ $$$${rejecting}\:+{ve}\:{sign} \\ $$$$\Rightarrow\:\:{x}={b}−\frac{{b}^{\mathrm{2}} }{{a}+{b}}\: \\ $$$$\:\:\:\:\:\:\:{x}=\frac{{ab}}{{a}+{b}}\:\:\:. \\ $$

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