Difference of Roots

We have the equation: $$12x^{2}-20x+3\lambda=0, \lambda\in Z$$

Sum of the roots: $$\alpha+\beta\ =-\dfrac{b}{a}=-\ \dfrac{\left(-20\right)}{12}=\dfrac{5}{3}$$

Product of the roots: $$\alpha\ \beta\ =\dfrac{c}{a}=\dfrac{3λ}{12}=\dfrac{λ}{4}$$

Now, we know that: $$\left(\beta\ -\alpha\ \right)^2=\left(\beta\ +\alpha\ \right)^2-4\alpha\ \beta\ $$

$$\left(\beta\ -\alpha\ \right)^2=\left(\dfrac{5}{3}\right)^2-4\left(\dfrac{λ}{4}\right)$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25}{9}-λ$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25-9λ}{9}$$

Taking the square root of both sides:

$$\left|\beta\ -\alpha\right|=\dfrac{\sqrt{\left(25-9λ\right)}}{3}$$

Now, we are given that: 

$$\dfrac{1}{2}\le\ \left|\beta\ -\alpha\right|\le\ \dfrac{3}{2}$$

Substituting:

$$\dfrac{1}{2}\le\ \dfrac{\sqrt{\left(25-9λ\right)}}{3}\le\dfrac{3}{2}$$

Multiplying by 3:

$$\dfrac{3}{2}\le\ \sqrt{\left(25-9λ\right)}\le\dfrac{9}{2}$$

Or, $$\dfrac{9}{4}\le\ 25-9λ\le\dfrac{81}{4}$$

Subtracting 25 from all three:

$$-\dfrac{91}{4}\le\ -9λ\le-\dfrac{19}{4}$$

Dividing all three by -9:

$$\dfrac{91}{36}\ge\ \ λ\ge\dfrac{19}{36}$$

Or, we can say, $$0.527 \leq \lambda \leq 2.527$$

The only integers that fall within the range [0.527, 2.527] are: 1 and 2. 

Hence, the sum of all possible values of $$\lambda$$ is 3.

Given, $$ \lambda x^2 - (\lambda+3)x + 3 = 0 $$ with roots $$ \alpha < \beta$$ , and $$ \dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac{1}{3} $$

Sum of the roots, $$ \alpha + \beta = \dfrac{\lambda+3}{\lambda}, \quad \alpha\beta = \dfrac{3}{\lambda} $$

Now, $$ \dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac{\beta - \alpha}{\alpha\beta} $$

So, $$ \dfrac{\beta - \alpha}{\alpha\beta} = \dfrac{1}{3} $$

$$  \Rightarrow \beta - \alpha = \dfrac{\alpha\beta}{3} = \dfrac{1}{\lambda} $$

Now using, $$ (\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta $$

$$ \left(\dfrac{1}{\lambda}\right)^2 = \left(\dfrac{\lambda+3}{\lambda}\right)^2 - \dfrac{12}{\lambda} $$

or, $$ 1 = (\lambda+3)^2 - 12\lambda $$

or, $$ 1 = \lambda^2 + 6\lambda + 9 - 12\lambda = \lambda^2 - 6\lambda + 9 $$

or, $$ \lambda^2 - 6\lambda + 8 = 0$$

or, $$ (\lambda - 2)(\lambda - 4) = 0$$

or, $$ \lambda = 2, 4 $$

On putting the values and checking the discriminants, we observe that both give real and distinct roots, so both are valid.

So, the sum of all the values of $$\lambda$$ is $$2 + 4 = \boxed{6} $$

We have $$f(x) = (x - p)^2 - q$$, $$q > 0$$, and $$a_1, a_2, a_3, a_4$$ in AP with mean $$p$$ and positive common difference $$d$$.

Since the four terms are in AP with mean $$p$$, we have $$\frac{a_1 + a_2 + a_3 + a_4}{4} = p$$.

We may therefore set $$a_1 = p - \frac{3d}{2}$$, $$a_2 = p - \frac{d}{2}$$, $$a_3 = p + \frac{d}{2}$$, and $$a_4 = p + \frac{3d}{2}$$.

Substituting into $$f$$ gives $$f(a_1) = \left(\frac{3d}{2}\right)^2 - q = \frac{9d^2}{4} - q$$, $$f(a_2) = \left(\frac{d}{2}\right)^2 - q = \frac{d^2}{4} - q$$, $$f(a_3) = \left(\frac{d}{2}\right)^2 - q = \frac{d^2}{4} - q$$, and $$f(a_4) = \left(\frac{3d}{2}\right)^2 - q = \frac{9d^2}{4} - q$$.

Since each of these values has absolute value $$500$$, we obtain $$\left|\frac{9d^2}{4} - q\right| = 500$$ and $$\left|\frac{d^2}{4} - q\right| = 500$$.

Because $$q > 0$$ and the two expressions must both have absolute value $$500$$ with different inputs, they must have opposite signs (otherwise $$\frac{9d^2}{4} - q = \frac{d^2}{4} - q$$ would force $$d = 0$$, contradicting the positive common difference).

Thus we consider the case $$\frac{9d^2}{4} - q = 500$$ and $$\frac{d^2}{4} - q = -500$$.

Subtracting the second from the first yields $$\frac{9d^2}{4} - \frac{d^2}{4} = 1000$$, which simplifies to $$2d^2 = 1000$$ and therefore $$d^2 = 500$$.

Substituting into the second equation gives $$q = \frac{d^2}{4} + 500 = 125 + 500 = 625$$.

Hence $$(x - p)^2 = q = 625$$, so $$x - p = \pm 25$$ and the two roots are $$p + 25$$ and $$p - 25$$.

The absolute difference between the roots is $$50$$, and therefore the answer is 50.