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# Help!!!!!

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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|?

Sep 25, 2017

#1
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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

Sep 25, 2017
edited by Melody  Sep 25, 2017

#1
+2

$$x = {1+\dfrac{\sqrt{2}}{1+\dfrac{\sqrt{2}}{1+...}}}$$.

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

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

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