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.
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.
On
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 $t=20x$ $$t^2-6t-40=0$$ $$(t-10)(t+4)=0$$ $$(20x-10)(20x+4)=0$$ $$\dots$$
HINT: by the quadradic formula we get $$x_{1,2}=\frac{3}{20}\pm \sqrt{\frac{9}{400}+\frac{40}{400}}$$