Mathematica

Algebra Level 3

x + 1 + x − 1 x + 1 − x − 1 = 4 x − 1 2 \frac { \sqrt { x+1 } +\sqrt { x-1 } }{ \sqrt { x+1 } -\sqrt { x-1 } } =\frac { 4x-1 }{ 2 }

Find the value of x x satisfying the above equation. Enter your answer rounded to two decimal places.


The answer is 1.25.

Syed Baqir by Syed Baqir , 24, Cyprus

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2 solutions

Tapas Mazumdar
Mar 23, 2017

x + 1 + x − 1 x + 1 − x − 1 = 4 x − 1 2 ⟹ x + 1 + x − 1 x + 1 − x − 1 ⋅ x + 1 + x − 1 x + 1 + x − 1 = 4 x − 1 2 ⟹ 2 x + 2 x 2 − 1 2 = 4 x − 1 2 ⟹ 2 x + 2 x 2 − 1 = 4 x − 1 ⟹ 2 x 2 − 1 = 2 x − 1 ⟹ ( 2 x 2 − 1 ) 2 = ( 2 x − 1 ) 2 ⟹ 4 x 2 − 4 = 4 x 2 − 4 x + 1 ⟹ 4 x = 5 ⟹ x = 5 4 = 1.25 \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \dfrac{4x-1}{2} \\ \implies \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \cdot \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} = \dfrac{4x-1}{2} \\ \implies \dfrac{2x + 2 \sqrt{x^2-1}}{2} = \dfrac{4x-1}{2} \\ \implies 2x + 2 \sqrt{x^2-1} = 4x-1 \\ \implies 2 \sqrt{x^2-1} = 2x-1 \\ \implies {\left( 2 \sqrt{x^2-1} \right)}^2 = {(2x-1)}^2 \\ \implies 4x^2 - 4 = 4x^2 - 4x +1 \\ \implies 4x = 5 \\ \implies x = \dfrac 54 = \boxed{1.25}

2 Helpful 1 Interesting 1 Brilliant 0 Confused
Syed Baqir
Aug 25, 2015

using C o m p e n d e n d o a n d D i v i d e n d o W e h a v e , x + 1 x − 1 = 4 x + 1 4 x − 3 ∴ x + 1 x − 1 = ( 4 x + 1 ) 2 ( 4 x − 3 ) 2 ⟹ x + 1 x − 1 = 16 x 2 + 1 + 8 x 16 x 2 + 9 − 24 x A g a i n C o m p e n d e n d o a n d D i v i d e n d o , M u l t i p l y i n g 2 a n d d i v i d i n g 2 2 2 ∗ x 1 = 32 x 2 − 16 x + 10 32 x − 8 ⟹ 16 x 2 − 4 x = 16 x 2 − 8 x + 5 ∴ x = 5 4 \quad Compendendo\quad and\quad Dividendo\quad \\ We\quad have,\quad \\ \frac { \sqrt { x+1 } }{ \sqrt { x-1 } } =\frac { 4x+1 }{ 4x-3 } \\ \therefore \quad \frac { x+1 }{ x-1 } =\frac { (4x+1)^{ 2 } }{ { (4x-3) }^{ 2 } } \Longrightarrow \quad \frac { x+1 }{ x-1 } =\frac { 16{ x }^{ 2 }+1+8x }{ 16{ x }^{ 2 }+9-24x } \\ Again\quad Compendendo\quad and\quad Dividendo,\\ Multiplying\quad 2\quad and\quad dividing\quad 2\\ \frac { 2 }{ 2 } *\frac { x }{ 1 } =\frac { 32{ x }^{ 2 }-16x+10 }{ 32x-8 } \Longrightarrow 16{ x }^{ 2 }-4x\quad =\quad 16{ x }^{ 2 }-8x\quad +5\\ \therefore \quad x\quad =\quad \frac { 5 }{ 4 }

2 Helpful 1 Interesting 0 Brilliant 0 Confused

C o m p e n d e n d o a n d D i v i d e n d o t h e o r e m i s a s f o l l o w s I f y o u h a v e : a : b = c : d t h e n , a + b : a − b = c + d : c − d \quad Compendendo\quad and\quad Dividendo\quad theorem\quad is\quad as\quad follows\\ If\quad you\quad have:\quad a:b=c:d\\ then,\quad \quad \quad \quad \quad \quad \quad \quad a+b:a-b=c+d:c-d\quad

Syed Baqir - 5 years, 9 months ago

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Can you prove that theorem?

Ben Habeahan - 5 years, 9 months ago

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a b = c d = k a = b k , c = d k a + b a − b = c + d c − d ⟹ b ( k + 1 ) b ( k − 1 ) = d ( k + 1 ) d ( k − 1 ) ( k + 1 ) ( k − 1 ) = ( k + 1 ) ( k − 1 ) 1 = 1 H e n c e P r o v e d \frac { a }{ b } =\frac { c }{ d } =k\\ a\quad =\quad bk\quad ,\quad c\quad =\quad dk\\ \frac { a+b }{ a-b } =\frac { c+d }{ c-d } \Longrightarrow \frac { b(k+1) }{ b(k-1) } =\frac { d(k+1) }{ d(k-1) } \\ (k+1)(k-1)\quad =\quad (k+1)(k-1)\\ 1\quad =\quad 1\\ Hence\quad Proved

Syed Baqir - 5 years, 9 months ago

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@Syed Baqir – Owww..thanks for the prove

Ben Habeahan - 5 years, 9 months ago

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@Ben Habeahan – np , I am glad it helped !

Syed Baqir - 5 years, 9 months ago

@Ben Habeahan – Check now I have edited Latex

Syed Baqir - 5 years, 9 months ago

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