<s>[INST] How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confused there.
 [/INST] Since the roots are $4/5\pm\sqrt{31}/5$, we have that $x^2+(a/5)x+b/5=(x-(4/5+\sqrt{31}/5))(x-(4/5-\sqrt{31}/5))$. 
So, $x^2+(a/5)x+b/5=x^2-(8/5)x+(16/25-31/25)=x^2-(8/5)x-3/5$.
Hence $a=-8, b=-3$.
</s>