(√(x^2 +ax−1))−(√(x^2 +bx−1))=(√a)−(√b) find x .


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Question Number 192777 by York12 last updated on 26/May/23

$\sqrt{x^{2} + a x - 1} - \sqrt{x^{2} + b x - 1} = \sqrt{a} - \sqrt{b}$ $f i n d x .$
Commented by Frix last updated on 27/May/23

$Obviously x = 1$ $The 2^{nd} solution after squaring, transforming$ $etc . is$ $x = \frac{a + b + 4 - 2 \sqrt{a b}}{a + b - 4 + 2 \sqrt{a b}}$
Commented by York12 last updated on 27/May/23
$x((√(a )) + (√b)) = (√(x^2 +ax−1)) + (√(x^2 +bx−1)) .....(i) (√a) − (√b) = (√(x^2 +ax−1)) − (√(x^2 +bx−1)) ......(ii) By multiplying (i) and (ii) we get : (x^2 +ax−1) − (x^2 +bx−1) = x (a−b) = (a−b) x =1 →(I) By adding (i) and (ii) we get : 2(√(x^2 +ax−1)) = ((√a)−(√b))x+((√a)+(√b)) 4x^2 + 4ax −4 = ((√a)−(√b))x^2 +2(a−b)x+((√a)+(√b))^2 By transforming and solving the quadratic in x we get x= ((((√a)−(√b))^2 +4)/(((√a)+(√b))^2 −4)) → (II) By (I) and (II) x ∈ { 1 , ((((√a)−(√b))^2 +4)/(((√a)+(√b))^2 −4)) }$
$x (\sqrt{a} + \sqrt{b}) = \sqrt{x^{2} + a x - 1} + \sqrt{x^{2} + b x - 1} . . . . . (i)$ $\sqrt{a} - \sqrt{b} = \sqrt{x^{2} + a x - 1} - \sqrt{x^{2} + b x - 1} . . . . . . (i i)$ $B y m u l t i p l y i n g (i) a n d (i i) w e g e t :$ $(x^{2} + a x - 1) - (x^{2} + b x - 1) = x (a - b) = (a - b)$ $x = 1 \to (I)$ $B y a d d i n g (i) a n d (i i) w e g e t :$ $2 \sqrt{x^{2} + a x - 1} = (\sqrt{a} - \sqrt{b}) x + (\sqrt{a} + \sqrt{b})$ $4 x^{2} + 4 a x - 4 = (\sqrt{a} - \sqrt{b}) x^{2} + 2 (a - b) x + {(\sqrt{a} + \sqrt{b})}^{2}$ $B y t r a n s f o r m i n g a n d s o l v i n g t h e q u a d r a t i c i n x$ $w e g e t$ $x = \frac{{(\sqrt{a} - \sqrt{b})}^{2} + 4}{{(\sqrt{a} + \sqrt{b})}^{2} - 4} \to (I I)$ $B y (I) a n d (I I) x \in {1, \frac{{(\sqrt{a} - \sqrt{b})}^{2} + 4}{{(\sqrt{a} + \sqrt{b})}^{2} - 4}}$
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