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# Given x^2-3x-1=0, what is 1/x-x ?

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Given $x^2-3x-1=0,$ what is $\dfrac{1}{x}-x ?$

Thanks for helping!

Nov 4, 2020

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Add 1 to both sides: $x^2-3x=1$

Divide by x: $x-3 = \frac{1}{x}$

Subtract x from both sides: $-3 = \frac{1}{x} - x$

Our answer is $$-3$$

Nov 4, 2020