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\(x = {1+\frac{\sqrt{2}}{1+\frac{\sqrt{2}}{1+...}}}\). Find \(\frac{1}{(x+1)(x-2)}\). When your answer is in the form \(\frac{A+\sqrt{B}}{C}\) , where A, B, and C are integers, what is |A|+|B|+|C|?

MIRB16  Sep 25, 2017

Best Answer 

 #1
avatar+92624 
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\(x = {1+\dfrac{\sqrt{2}}{1+\dfrac{\sqrt{2}}{1+...}}}\).

 

 

 

\(x=1+\frac{\sqrt2}{x}\\ x^2=x+\sqrt2\\ x^2-x=\sqrt2\\ \)

 

 

    \(\frac{1}{(x+1)(x-2)}\\ =\frac{1}{(x+1)(x-2)}\\ =\frac{1}{x^2-x-2}\\ =\frac{1}{\sqrt2-2}\\ =\frac{-1}{2-\sqrt2}\\ =\frac{-1}{2-\sqrt2}\times \frac{2+\sqrt2}{2+\sqrt2}\\ =\frac{-(2+\sqrt2)}{4-2}\\ =\frac{-2-\sqrt2}{2}\\ =\frac{2+\sqrt2}{-2}\\ \)  .

 

 

When your answer is in the form   \(\frac{A+\sqrt{B}}{C}\)   ,

 

A=+2,   B=+2,   C= -2

 

where A, B, and C are integers, what is |A|+|B|+|C|?

 

|A|+|B|+|C| = 2+2+2 = 6

Melody  Sep 25, 2017
edited by Melody  Sep 25, 2017
 #1
avatar+92624 
+2
Best Answer

\(x = {1+\dfrac{\sqrt{2}}{1+\dfrac{\sqrt{2}}{1+...}}}\).

 

 

 

\(x=1+\frac{\sqrt2}{x}\\ x^2=x+\sqrt2\\ x^2-x=\sqrt2\\ \)

 

 

    \(\frac{1}{(x+1)(x-2)}\\ =\frac{1}{(x+1)(x-2)}\\ =\frac{1}{x^2-x-2}\\ =\frac{1}{\sqrt2-2}\\ =\frac{-1}{2-\sqrt2}\\ =\frac{-1}{2-\sqrt2}\times \frac{2+\sqrt2}{2+\sqrt2}\\ =\frac{-(2+\sqrt2)}{4-2}\\ =\frac{-2-\sqrt2}{2}\\ =\frac{2+\sqrt2}{-2}\\ \)  .

 

 

When your answer is in the form   \(\frac{A+\sqrt{B}}{C}\)   ,

 

A=+2,   B=+2,   C= -2

 

where A, B, and C are integers, what is |A|+|B|+|C|?

 

|A|+|B|+|C| = 2+2+2 = 6

Melody  Sep 25, 2017
edited by Melody  Sep 25, 2017

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