Al-Khwarizmi, the father of algebra

Algebra Level 3

If 1 + x + 1 − x 1 + x − 1 − x = R \dfrac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=R , then what is the value of 2 R 2 − 4 R x 2R^{2}-\dfrac{4R}{x} ?


The answer is -2.

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Nazmus Sakib by Nazmus sakib , 24, Bangladesh

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4 solutions

Chew-Seong Cheong
Jan 14, 2018

R = 1 + x + 1 − x 1 + x − 1 − x Multiplying up and down by 1 + x + 1 − x = ( 1 + x + 1 − x ) 2 ( 1 + x ) 2 − ( 1 − x ) 2 = 1 + 1 − x 2 x Rearranging R x − 1 = 1 − x 2 Squaring both sides R 2 x 2 − 2 R x + 1 = 1 − x 2 R 2 x 2 − 2 R x = − x 2 Multiplying both sides by 2 x 2 2 R − 4 R x = − 2 \begin{aligned} R & = \color{#3D99F6} \frac {\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} & \small \color{#3D99F6} \text{Multiplying up and down by }{\sqrt{1+x}+\sqrt{1-x}} \\ & = \color{#3D99F6} \frac {\left(\sqrt{1+x}+\sqrt{1-x}\right)^2}{\left(\sqrt{1+x}\right)^2-\left(\sqrt{1-x}\right)^2} \\ & = \frac {1+\sqrt{1-x^2}}x & \small \color{#3D99F6} \text{Rearranging} \\ Rx - 1 & = \sqrt{1-x^2} & \small \color{#3D99F6} \text{Squaring both sides} \\ R^2x^2 - 2Rx + 1 & = 1-x^2 \\ R^2x^2 - 2Rx & = -x^2 & \small \color{#3D99F6} \text{Multiplying both sides by }\frac 2{x^2} \\ 2R - \frac {4R}x & = \boxed{-2} \end{aligned}

8 Helpful 0 Interesting 0 Brilliant 0 Confused
Abhinav Jain
Jan 15, 2018

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Munem Shahriar
Jan 14, 2018

Relevant wiki: Componendo and Dividendo

1 + x + 1 − x 1 + x − 1 − x = R \dfrac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=R

⇒ ( 1 + x + 1 − x ) + ( 1 + x − 1 − x ) ( 1 + x + 1 − x ) − ( 1 + x − 1 − x ) = R + 1 R − 1 \Rightarrow \dfrac{(\sqrt{1+x}+\sqrt{1-x}) + (\sqrt{1+x}-\sqrt{1-x})}{(\sqrt{1+x}+\sqrt{1-x}) - (\sqrt{1+x}-\sqrt{1-x})} = \dfrac{R+1}{R-1}

⇒ 1 + x + 1 + x 1 − x + 1 − x = R + 1 R − 1 \Rightarrow \dfrac{\sqrt{1+x}+ \sqrt{1+x}}{\sqrt{1-x}+\sqrt{1-x}} = \dfrac{R+1}{R-1}

⇒ 2 1 + x 2 1 − x = R + 1 R − 1 \Rightarrow\dfrac{2\sqrt{1+x}}{2\sqrt{1-x}} = \dfrac{R+1}{R-1}

⇒ 1 + x 1 − x = R + 1 R − 1 \Rightarrow \dfrac{\sqrt{1+x}}{\sqrt{1-x}} = \dfrac{R+1}{R-1}

⇒ ( 1 + x ) 2 ( 1 − x ) 2 = ( R + 1 ) 2 ( R − 1 ) 2 \Rightarrow\dfrac{(\sqrt{1+x})^2}{(\sqrt{1-x})^2} =\dfrac{(R+1)^2}{(R-1)^2}

⇒ 1 + x 1 − x = R 2 + 2 R + 1 2 R 2 − 2 R + 1 2 \Rightarrow\dfrac{1+x}{1-x} = \dfrac{R^2+2R+1^2}{R^2-2R+1^2}

⇒ 1 + x + 1 − x 1 + x − 1 + x = R 2 + 2 R + 1 2 + R 2 − 2 R + 1 2 R 2 + 2 R + 1 2 − R 2 + 2 R − 1 2 \Rightarrow\dfrac{1+x+1-x}{1+x-1+x} = \dfrac{R^2+2R+1^2 +R^2-2R+1^2 }{R^2+2R+1^2 - R^2+2R-1^2 }

⇒ 2 2 x = 2 R 2 + 2 4 R \Rightarrow\dfrac{2}{2x} = \dfrac{2R^2 + 2}{4R}

⇒ 1 x = 2 ( R 2 + 1 ) 4 R \Rightarrow\dfrac1x = \dfrac{2(R^2+1)}{4R}

⇒ 1 x = R 2 + 1 2 R \Rightarrow\dfrac 1x = \dfrac{R^2 + 1}{2R}

⇒ 2 R x = R 2 + 1 \Rightarrow\dfrac{2R}{x} = R^2+1

⇒ 2 ( 2 R x ) = 2 ( R 2 + 1 ) \Rightarrow2\left(\dfrac{2R}{x}\right) = 2(R^2+1)

⇒ 4 R x = 2 R 2 + 2 \Rightarrow\dfrac{4R}{x} = 2R^2+2

⟹ − 2 = 2 R 2 − 4 R x \implies -2 = 2R^2 - \dfrac{4R}{x}

Hence 2 R 2 − 4 R x = − 2 2R^2 - \dfrac{4R}{x} = \boxed{-2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

In this step: 1 + x + 1 − x 1 + x − 1 + x = R 2 + 2 R + 1 2 + R 2 − 2 R + 1 2 R 2 + 2 R + 1 2 − R 2 + 2 R − 1 2 \dfrac{1+x+1-x}{1+x-1+x} = \dfrac{R^2+2R+1^2 +R^2-2R+1^2 }{R^2+2R+1^2 - R^2+2R-1^2 } , I think there is a mistake in the denominator of RHS based on the previous step in your solution.

Anuj Shikarkhane - 2 years, 12 months ago
Geoff Taylor
Jan 15, 2018

Using a direct approach R = ( 1 + x + 1 − x ) ( 1 + x + 1 − x ) 2 x R=\frac { \left( \sqrt { 1+x } +\sqrt { 1-x } \right) (\sqrt { 1+x } +\sqrt { 1-x } ) }{ 2x } after rationalizing the denominator simplifies to R = 1 + 1 − x 2 x R=\frac { 1+\sqrt { 1-{ x }^{ 2 } } }{ x }

R 2 = ( 1 + 1 − x 2 x ) 2 = 2 + 2 1 − x 2 − x 2 x 2 { R }^{ 2 }={ \left( \frac { 1+\sqrt { 1-{ x }^{ 2 } } }{ x } \right) }^{ 2 }=\frac { 2+2\sqrt { 1-{ x }^{ 2 } } -{ x }^{ 2 } }{ { x }^{ 2 } }

so 2 R 2 − 4 R x 2{ R }^{ 2 }-\frac { 4R }{ x }

= 2 ( 2 + 2 1 − x 2 − x 2 ) − 4 ( 1 + 1 − x 2 ) x 2 = − 2 \frac { 2\left( 2+2\sqrt { 1-{ x }^{ 2 } } -{ x }^{ 2 } \right) -4\left( 1+\sqrt { 1-{ x }^{ 2 } } \right) }{ { x }^{ 2 } } =\boxed { -2 }

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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