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Question Number 192777 by York12 last updated on 26/May/23
(√(x^2 +ax−1))−(√(x^2 +bx−1))=(√a)−(√b) find x .
$$\sqrt{{x}^{\mathrm{2}} +{ax}−\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +{bx}−\mathrm{1}}=\sqrt{{a}}−\sqrt{{b}} \\ $$$${find}\:{x}\:. \\ $$
Commented by Frix last updated on 27/May/23
Obviously x=1 The 2^(nd) solution after squaring, transforming etc. is x=((a+b+4−2(√(ab)))/(a+b−4+2(√(ab))))
$$\mathrm{Obviously}\:{x}=\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution}\:\mathrm{after}\:\mathrm{squaring},\:\mathrm{transforming} \\ $$$$\mathrm{etc}.\:\mathrm{is} \\ $$$${x}=\frac{{a}+{b}+\mathrm{4}−\mathrm{2}\sqrt{{ab}}}{{a}+{b}−\mathrm{4}+\mathrm{2}\sqrt{{ab}}} \\ $$
Commented by York12 last updated on 27/May/23
x((√(a )) + (√b)) = (√(x^2 +ax−1)) + (√(x^2 +bx−1)) .....(i) (√a) − (√b) = (√(x^2 +ax−1)) − (√(x^2 +bx−1)) ......(ii) By multiplying (i) and (ii) we get : (x^2 +ax−1) − (x^2 +bx−1) = x (a−b) = (a−b) x =1 →(I) By adding (i) and (ii) we get : 2(√(x^2 +ax−1)) = ((√a)−(√b))x+((√a)+(√b)) 4x^2 + 4ax −4 = ((√a)−(√b))x^2 +2(a−b)x+((√a)+(√b))^2 By transforming and solving the quadratic in x we get x= ((((√a)−(√b))^2 +4)/(((√a)+(√b))^2 −4)) → (II) By (I) and (II) x ∈ { 1 , ((((√a)−(√b))^2 +4)/(((√a)+(√b))^2 −4)) }
$$ \\ $$$${x}\left(\sqrt{{a}\:}\:+\:\sqrt{{b}}\right)\:=\:\sqrt{{x}^{\mathrm{2}} +{ax}−\mathrm{1}}\:\:+\:\sqrt{{x}^{\mathrm{2}} +{bx}−\mathrm{1}}\:\:…..\left({i}\right) \\ $$$$\sqrt{{a}}\:−\:\sqrt{{b}}\:=\:\sqrt{{x}^{\mathrm{2}} +{ax}−\mathrm{1}}\:\:−\:\sqrt{{x}^{\mathrm{2}} +{bx}−\mathrm{1}}\:\:\:\:……\left({ii}\right) \\ $$$${By}\:{multiplying}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}\:: \\ $$$$\left({x}^{\mathrm{2}} +{ax}−\mathrm{1}\right)\:−\:\left({x}^{\mathrm{2}} +{bx}−\mathrm{1}\right)\:=\:{x}\:\left({a}−{b}\right)\:=\:\left({a}−{b}\right) \\ $$$${x}\:=\mathrm{1}\:\rightarrow\left({I}\right) \\ $$$${By}\:{adding}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}\:: \\ $$$$\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{ax}−\mathrm{1}}\:=\:\left(\sqrt{{a}}−\sqrt{{b}}\right){x}+\left(\sqrt{{a}}+\sqrt{{b}}\right) \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{4}{ax}\:−\mathrm{4}\:=\:\left(\sqrt{{a}}−\sqrt{{b}}\right){x}^{\mathrm{2}} +\mathrm{2}\left({a}−{b}\right){x}+\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} \\ $$$${By}\:{transforming}\:{and}\:{solving}\:{the}\:{quadratic}\:{in}\:{x} \\ $$$${we}\:{get}\: \\ $$$${x}=\:\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} +\mathrm{4}}{\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} −\mathrm{4}}\:\rightarrow\:\left({II}\right)\: \\ $$$${By}\:\left({I}\right)\:{and}\:\left({II}\right)\:{x}\:\in\:\left\{\:\mathrm{1}\:,\:\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} +\mathrm{4}}{\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} −\mathrm{4}}\:\right\} \\ $$

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