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# help

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If x^2 + 1/x^2 = 171, what is the positive value of x - 1/x?

Jan 11, 2020

#1
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$$x^2+\frac{1}{x^2}=171$$

$$x-\frac{1}{x}$$ let's square this

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

we know that $$x^2+\frac{1}{x^2}=171$$

so $$(x-\frac{1}{x})^2=169$$

To find x-1/x we square root then it is =13

Jan 11, 2020
edited by Guest  Jan 11, 2020
edited by Guest  Jan 11, 2020
#2
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If x^2 + 1/x^2 = 171, what is the positive value of x - 1/x?

Hello Guest!

$$x^2+\frac{1}{x^2}=171\\ \frac{x^4+1}{x^2}=171\\x^4+1=171x^2$$

$$\color{blue}x^2=a\\ a^2+1=171a\\ a^2-171a+1=0$$

$$a=\frac{171}{2}\pm\sqrt{(\frac{171}{2})^2-1}$$

$$a_1=170.99415\\ x_{1,1}=13,0764732\\ x_{1,2}=-13,0764732\\ a_2=0.0058481532\\ x_{2,1}=0.0764732\\ x_{2,2}=-0,0764732$$

$$x_{1,1}-\frac{1}{x_{1,1}}=13,0764732-\frac{1}{13,0764732}\color{blue} =13\\ x_{1,2}-\frac{1}{x_{1,2}}=-13,0764732-\frac{1}{-13,0764732}=-13\\ x_{2,1}-\frac{1}{x_{2,1}}=0.0764732-\frac{1}{0.0764732}=-13\\ x_{2,2}-\frac{1}{x_{2,2}}=-0.0764732-\frac{1}{-0.0764732}\color{blue}=13\\$$

The positive value of x - 1/x is 13.

!

Jan 11, 2020