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Algebra Level 4

x 2 + 1 − 1 x 2 − 5 3 = x \large \sqrt{x^2+1} - \frac { 1} { \sqrt{x^2 - \frac 53} } = x

There is a unique solution of the form − a b - \dfrac{a}{b} , where a a and b b are coprime positive integers. Find a + b a + b .


The answer is 7.

Rakshit Joshi by Rakshit Joshi , 26, India

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2 solutions

x 2 + 1 = x + 1 x 2 − 5 3 . S q u a r i n g a n d s i m p l i f y i n g , 0 = 1 x 2 − 5 3 − 1 + 2 x x 2 − 5 3 . ∴ 8 3 − x 2 x 2 − 5 3 = 2 x . S q u a r i n g , x 4 − 16 3 x 2 + 64 9 x 2 − 5 3 = 4 x 2 , ∴ x 4 − 16 3 x 2 + 64 9 = 4 x 4 − 20 3 x 2 . S i m p l i f y 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 2 , 3 x 4 − 20 3 x 2 − 64 9 = 0. x 2 = 4 ± 28 18 . x 2 > 0 , ∴ x 2 = 16 9 . ⟹ x = ± 4 3 . B u t x = − 4 3 o n l y s a t i s f y t h e o r i g i n a l e q u a t i o n . \sqrt{x^2+1}=x+\dfrac1 {\sqrt{x^2-\frac 5 3}}. ~~~~~Squaring~and~simplifying,\\ 0=\dfrac1 {x^2-\frac 5 3} - 1+\dfrac{2x} {\sqrt{x^2-\frac 5 3}}.\\ \therefore~\dfrac{\frac8 3 - x^2} {\sqrt{x^2-\frac 5 3}}= 2x.~~~~~Squaring,\\ \dfrac{x^4 - \dfrac{16}3 x^2+\dfrac{64} 9}{x^2-\frac 5 3 }=4x^2,\\ \therefore~x^4 - \dfrac{16}3 x^2+\dfrac{64} 9=4x^4-\frac{20}3 x^2. \\ Simplifying~and ~solving~ the~ quadratic ~in~ x^2 , \\ 3x^4 - \frac{20}3 x^2 - \dfrac{64} 9=0. \\ x^2=\dfrac{4~\pm 28} {18}. \\ x^2>0,~\therefore~x^2=\dfrac{16} 9.\\ \implies~ x= \pm \frac 4 3.\\ But~~x= {\Large~~~\color{#D61F06}{ - \frac 4 3}}~only~satisfy ~the ~original~ equation.

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x 2 + 1 − 1 x 2 − 5 3 = x Let u = x 2 − 1 3 u + 4 3 − 1 u − 4 3 = u + 1 3 Multiply both sides by u − 4 3 u 2 − 16 9 − 1 = u 2 − u − 4 9 Squaring both sides u 2 − 16 9 − 2 u 2 − 16 9 + 1 = u 2 − u − 4 9 Rearranging u − 1 3 = 2 u 2 − 16 9 Multiply both sides by 3 3 u − 1 = 2 9 u 2 − 16 Squaring both sides 9 u 2 − 6 u + 1 = 36 u 2 − 64 Rearranging 27 u 2 + 6 u − 65 = 0 ( 9 u − 13 ) ( 3 u + 5 ) = 0 \begin{aligned} \sqrt{x^2+1} - \frac 1{\sqrt{x^2-\frac 53}} & = x & \small \color{#3D99F6} \text{Let } u = x^2 - \frac 13 \\ \sqrt{u+\frac 43} - \frac 1{\sqrt{u-\frac 43}} & = \sqrt{u+\frac 13} & \small \color{#3D99F6} \text{Multiply both sides by } \sqrt{u - \frac 43} \\ \sqrt{u^2 - \frac {16}9} - 1 & = \sqrt{u^2-u - \frac 49} & \small \color{#3D99F6} \text{Squaring both sides} \\ u^2 - \frac {16}9 - 2 \sqrt{u^2 - \frac {16}9} + 1 & = u^2- u - \frac 49 & \small \color{#3D99F6} \text{Rearranging} \\ u - \frac 13 & = 2 \sqrt{u^2 - \frac {16}9} & \small \color{#3D99F6} \text{Multiply both sides by }3 \\ 3u - 1 & = 2 \sqrt{9u^2 - 16} & \small \color{#3D99F6} \text{Squaring both sides} \\ 9u^2 - 6u + 1 & = 36u^2 - 64 & \small \color{#3D99F6} \text{Rearranging} \\ 27u^2 + 6u - 65 & = 0 \\ (9u-13)(3u+5) & = 0 \end{aligned}

⟹ { u = 13 9 ⟹ x 2 = 13 9 + 1 3 = 16 9 ⟹ x = { 4 3 solution not valid − 4 3 solution valid u = − 5 3 ⟹ x 2 = − 5 3 + 1 3 = − 4 3 No real solution. \implies \begin{cases} u = \dfrac {13}9 & \implies x^2 = \dfrac {13}9 + \dfrac 13 = \dfrac {16}9 & \implies x = \begin{cases} \color{#D61F06} \frac 43 & \color{#D61F06} \small \text{ solution not valid} \\ \color{#3D99F6} - \frac 43 & \color{#3D99F6} \small \text{ solution valid} \end{cases} \\ u = - \dfrac 53 & \implies x^2 = - \dfrac 53 + \dfrac 13 = - \dfrac 43 & \small \color{#D61F06} \text{No real solution.} \end{cases}

Therefore a + b = 4 + 3 = 7 a+b=4+3 = \boxed 7 .

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