\(x^2+\frac{1}{x^2}=171\)
\(x-\frac{1}{x}\) 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
+14903 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.
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