We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$ 
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.

First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$
Can someone help me with the second one?
I forgot to tell that solving the equation is not an option in my case.
 A: Take the step below,
$$(x_1-x_2)^2= x_1^2+x_2^2 - 2x_1x_2=\dfrac{7}{4}-2(-\frac 12) = \frac{11}{4}
$$
Thus, 
$$|x_1-x_2|=\frac{\sqrt{11}}{2}$$
A: Well, you know that, by the Quadratic Formula, $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, so the difference between the two roots is $\frac{1}{a}\sqrt{b^2-4ac}=\frac{1}{a}\sqrt{3+4(2)(1)}=\frac{1}{2}\sqrt{11}$
A: Hint: What is $\tfrac{-b+\sqrt{b^2-4ac}}{2a} - \frac{-b-\sqrt{b^2-4ac}}{2a}$ ?
A: Note that: if $a,b,c \in \mathbb{R}$ and $a\ne0$, if $ax^2+bx+c=0$, then $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$2x^2-\sqrt{3}x-1=0$,
solving, we get: $x_{1,2}=\frac{\sqrt{3}\pm\sqrt{3-4\cdot2\cdot(-1)}}{2\cdot2}=\frac{\sqrt{3}\pm\sqrt{11}}{4}$
The required expression, $x_1^2+x_2^2=(\frac{\sqrt{3}+\sqrt{11}}{4})^2+(\frac{\sqrt{3}-\sqrt{11}}{4})^2=\frac{3+2\sqrt{33}+11+3-2\sqrt{33}+11}{16}=\frac{7}{4}$.
The second required expression, $|x_1-x_2|=|\frac{\sqrt{3}+\sqrt{11}}{4}-\frac{\sqrt{3}-\sqrt{11}}{4}|=|\frac{\sqrt{3}+\sqrt{11}-\sqrt{3}+\sqrt{11}}{4}|=|\frac{\sqrt{11}}{2}|=\frac{\sqrt{11}}{2}$.
