For $p,q\in R$, consider the real valued function $f\left(x\right)={\left(x-p\right)}^2-q,x\in R$ and $q>0$. Let $a_1,a_2,a_3$ and $a_4$ be in an arithmetic progression with mean $p$ and positive common difference. If $\left|f\left(a_i\right)\right|=500$ for all $i=1,2,3,4$, then the absolute difference between the roots of $f\left(x\right)=0$ is

Given,

$f\left(x\right)=0\Rightarrow {\left(x-p\right)}^2-q=0$

So, roots are $p+\sqrt{q},p-\sqrt{q}$

Now absolute difference between roots will be $2\sqrt{q}$.

Now given $a_1,a_2,a_3,a_4$ are in $A.P$ and its mean is $p$

Now let $a_1,a_2,a_3,a_4$  be  $a_1=p-3d, a_2=p-d, a_3=p+d \& a_4=p+3d$

Now given $\left|f\left(a_i\right)\right|=500$

So, $\left|f\left(a_4\right)\right|=500$

$\Rightarrow \left|{\left(a_4-p\right)}^2-q\right|=500$

$\Rightarrow {\left(a_4-p\right)}^2-q=500$

$\Rightarrow 9d^2-q=500 ....\left(1\right)$

And using $\left|f\left(a_i\right)\right|=500 \forall i=1,2,3,4$

We get ${\left|f\left(a_4\right)\right|}^2={\left|f\left(a_3\right)\right|}^2$

$\Rightarrow {\left({\left(a_4-p\right)}^2-q\right)}^2={\left({\left(a_3-p\right)}^2-q\right)}^2$

$\Rightarrow 9d^2-q+d^2-q=0$

So, $2q=10d^2\Rightarrow q=5d^2$

$\Rightarrow d^2=\frac{q}{5}$

From equation $\left(1\right)$ we get,

$9\left(\frac{q}{5}\right)-q=500$

$\Rightarrow \frac{4q}{5}=500$

$\Rightarrow q=\frac{500\times 5}{4}$

Now absolute difference is $2\sqrt{q}=2\times \sqrt{\frac{500\times 5}{4}}=2\times \frac{50}{2}=50$