Just for fun!-IV

Algebra Level 3

x + 1 + x − 1 x + 1 − x − 1 = 4 x − 1 2 , x = ? \large \dfrac { \sqrt { x+1 } +\sqrt { x-1 } }{ \sqrt { x+1 } -\sqrt { x-1 } } =\dfrac { 4x-1 }{ 2 } \quad, \quad x= \ ?


The answer is 1.25.

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Swapnil Das by Swapnil Das , 20, India

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2 solutions

Ikkyu San
Aug 18, 2015

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 ( x + 1 ) 2 + 2 ( x + 1 ) ( x − 1 ) + ( x − 1 ) 2 ( x + 1 ) 2 − ( x − 1 ) 2 = 4 x − 1 2 x + 1 + 2 ( 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 4 ( x 2 − 1 ) = 4 x 2 − 4 x + 1 4 x 2 − 4 = 4 x 2 − 4 x + 1 4 x = 5 x = 5 4 = 1.25 \begin{aligned}\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=&\ \dfrac{4x-1}2\\\dfrac{(\sqrt{x+1}+\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1})}{(\sqrt{x+1}-\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1})}=&\ \dfrac{4x-1}2\\\dfrac{(\sqrt{x+1})^2+2(\sqrt{x+1})(\sqrt{x-1})+(\sqrt{x-1})^2}{(\sqrt{x+1})^2-(\sqrt{x-1})^2}=&\ \dfrac{4x-1}2\\\dfrac{x+1+2\sqrt{(x+1)(x-1)}+x-1}{x+1-(x-1)}=&\ \dfrac{4x-1}2\\\dfrac{2x+2\sqrt{x^2-1}}2=&\ \dfrac{4x-1}2\\2x+2\sqrt{x^2-1}=&\ 4x-1\\2\sqrt{x^2-1}=&\ 2x-1\\4(x^2-1)=&\ 4x^2-4x+1\\4x^2-4=&\ 4x^2-4x+1\\4x=&\ 5\\x=&\ \dfrac54=\boxed{1.25}\end{aligned}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

This is easier

Shivam Sabharwal - 5 years, 9 months ago

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Depends on how one thinks :)

Swapnil Das - 5 years, 9 months ago

Nice solution. Could be easily done by componendo dividendo.

Swapnil Das - 5 years, 10 months ago
Navu Katz
Aug 22, 2015

Componendo-Dividendo: if a b = c d \frac{a}{b} = \frac{c}{d} then a + b a − b = c + d c − d \frac{a+b}{a-b} = \frac{c+d}{c-d}

Using this result,

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}

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

Squaring both sides,

x + 1 x − 1 = ( 4 x + 1 ) 2 ( 4 x − 3 ) 2 \frac{x+1}{x-1} = \frac{(4x+1)^{2}}{(4x-3)^{2}}

x + 1 x − 1 = 16 x 2 + 8 x + 1 16 x 2 − 24 x + 9 \frac{x+1}{x-1} = \frac{16x^{2}+8x+1}{16x^{2}-24x+9}

Using componendo-dividendo again,

x = 32 x 2 − 16 x + 10 32 x − 8 x = \frac{32x^{2}-16x+10}{32x-8}

32 x 2 − 8 x = 32 x 2 − 16 x + 10 32x^{2}-8x = 32x^{2}-16x+10

8 x = 10 8x = 10

x = 1.25 x = 1.25

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice.

Swapnil Das - 5 years, 9 months ago

Can you tell us how Using Compendendo Divindendo makes 16x^2 + 8x + 1 to 32x^2 - 24x +9 and similarly the denominator

Syed Baqir - 5 years, 9 months ago

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