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Solve: $10x-\frac{1}{x}=3$

This is what I've tried:

After the first step,

$\frac{10x^2-1}{x}=3$

$\Rightarrow$ $10x^2-3x-1=0$

But, I am not able to factorize this.

Any help would be much appreciated.

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    $\begingroup$ You can always use the quadratic formula $\endgroup$ – Alan Oct 3 '15 at 10:33
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    $\begingroup$ $(5x+1)(2x-1)$. $\endgroup$ – David Mitra Oct 3 '15 at 10:42
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HINT: by the quadradic formula we get $$x_{1,2}=\frac{3}{20}\pm \sqrt{\frac{9}{400}+\frac{40}{400}}$$

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  • $\begingroup$ I don't know about the quadratic formula. Is there any other way to solve the sum or can you explain me the quadratic formula, sir? $\endgroup$ – Abhishekstudent Oct 3 '15 at 10:38
  • $\begingroup$ see here purplemath.com/modules/quadform.htm $\endgroup$ – Dr. Sonnhard Graubner Oct 3 '15 at 10:40
  • $\begingroup$ quadratic formula : $$x=\frac{-b+-\sqrt(b^2-4ac)}{2a}$$ for standard form of quadratic equation $$ax^2+bx+c=0$$ $\endgroup$ – Adesh Tamrakar Oct 3 '15 at 10:44
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Here is an easy way of solving $10x^2-3x-1=0$.

What we would like to do is to turn $10x^2$ into a perfect square, but just multiplying both side of the equation by $10$ will leave fractions later on (which I would like to avoid). So, instead, I'm going to multiply by $2^2\cdot10 = 40$

$$40(10x^2-3x-1)=40\cdot0$$ $$400x^2-3(40)x-40=0$$ $$20^2x^2-6(20)x-40=0$$ $$(20x)^2-6(20x)-40=0$$ let $t=20x$ $$t^2-6t-40=0$$ $$(t-10)(t+4)=0$$ $$(20x-10)(20x+4)=0$$ $$\dots$$

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