# Need help in solving $10x-\frac1x=3$

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.

• You can always use the quadratic formula – Alan Oct 3 '15 at 10:33
• $(5x+1)(2x-1)$. – David Mitra Oct 3 '15 at 10:42

HINT: by the quadradic formula we get $$x_{1,2}=\frac{3}{20}\pm \sqrt{\frac{9}{400}+\frac{40}{400}}$$
• quadratic formula : $$x=\frac{-b+-\sqrt(b^2-4ac)}{2a}$$ for standard form of quadratic equation $$ax^2+bx+c=0$$ – Adesh Tamrakar Oct 3 '15 at 10:44
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 $t=20x$ $$t^2-6t-40=0$$ $$(t-10)(t+4)=0$$ $$(20x-10)(20x+4)=0$$ $$\dots$$